Geometry Word Problems (page 1 of 6)

Sections: Introduction, Basic examples, Triangle formulas, Complex examples, The Box Problem & the Goat Problem, Max / Min problems

In order to solve geometric word problems, you will need to have memorized some geometric formulas for at least the basic shapes (circles, squares, right triangles, etc). You will usually need to figure out from the word problem which formula to use, and many times you will need more than one formula for one exercise. So make sure you have memorized any formulas that are used in the homework, because you may be expected to know them on the test.

Some problems are just straightforward applications of basic geometric formulae.

The radius of a circle is 3 centimeters. What is the circle's circumference?

The formula for the circumference C of a circle with radius r is:

C = 2(pi)r

...where "pi" (above) is of course the number approximately equal to 22 / 7 or 3.14159. They gave me the value of r and asked me for the value of C, so I'll just "plug-n-chug":

C = 2(pi)(3) = 6pi

Then, after re-checking the original exercise for the required units (so my answer will be complete):

the circumference is 6pi cm.

Note: Unless you are told to use one of the approximations for pi, or are told to round to some number of decimal places (from having used the "pi" button on your calculator), you are generally supposed to keep your answer in "exact" form, as shown above. If you're not sure if you should use the "pi" form or the decimal form, use both: "6pi cm, or about 18.85 cm".

A square has an area of sixteen square centimeters. What is the length of each of its sides?

The formula for the area A of a square with side-length s is:

A = s2

They gave me the area, so I'll plug this value into the area formula, and see where this leads:

16 = s2 4 = s

After re-reading the exercise to find the correct units, my answer is:

The length of each side is 4 centimeters.

Most geometry word problems are a bit more involved than the example above. For most exercises, you will be given at least two pieces of information, such as a statement about a square's perimeter and then a question about its area. To find the solution, you will need to know the equations related to the various pieces of information; you will then probably solve one of the equations for a useful bit of new information, and then plug the result into another of the equations.

In other words, geometry word problems often aren't simple one-step exercises like the one shown above. But if you take all the information that you've been given, write down any applicable formulas, try to find ways to relate the various pieces, and see where this leads, then you'll almost always end up with a valid answer.

A cube has a surface area of fifty-four square centimeters. What is the volume of the cube?

The formula for the volume V of a cube with edge-length e is:

V = e3

To find the volume, I need the edge-length. Can I use the surface-area information to get what I need? Let's see...

A cube has six sides, each of which is a square; and the edges of the cube's faces are the sides of those squares. The formula for the area of a square with side-length e is A = e2. There are six faces so there are six squares, and the cube's total surface area SA must be:

SA = 6e2

Plugging in the value they gave me, I get:

54 = 6e2 54 / 6 = (6e2) / 6 9 = e2 3 = e

Since the volume is the cube of the edge-length, and since the units on this cube are centimeters, then:

the volume is 27 cubic centimeters, or 27 cc, or 27 mL

(The common abbreviation for "cubic centimeters" is "cc's", as you've no doubt heard on medical TV dramas, and one cc is equal in volume to one milliliter. So all three versions of the answer above are equivalent.)     Copyright © Elizabeth Stapel 2000-2008 All Rights Reserved

A circle has an area of 49pi square units. What is the length of the circle's diameter?

The formula for the area A of a circle with radius r is:

A = (pi)r2

I know that radius r is half of the length of the diameter d, so:

49pi = (pi)r2 (49pi) / pi = [(pi)r2] / pi 49 = r2 7 = r

Then the radius r has a length of 7 units, and:

the length of the diameter is 14 units

Warning: You can not assume that you will always be given all the geometric formulae on your tests. At some point, you will need to know at least some of them "by heart". The basic formulae you should know include the formulae for the area and perimeter (or circumference) of squares, rectangles, triangles, and circles; and the surface areas and volumes of cubes, rectangular solids, spheres, and cylinders. Depending on the class, you may also need to know the formulae for cones and pyramids. If you're not sure what your instructor expects, ask now.

Geometry Word Problems: Basic Examples(page 2 of 6)

Sections: Introduction, Basic examples, Triangle formulas, Complex examples, The Box Problem & the Goat Problem, Max / Min problems

Sometimes geometry word problems wrap the geometry in a thick layer of "real life". You will need to be able to "see" the geometry, and extract the relevant information.

Suppose a water tank in the shape of a right circular cylinder is thirty feet long and eight feet in diameter. How much sheet metal was used in its construction?

What they are asking for here is the surface area of the water tank. The total surface area of the tank will be the sum of the surface areas of the side (the cylindrical part) and of the ends. If the diameter is eight feet, then the radius is four feet. The surface area of each

end is given by the area formula for a circle with radius r: A = (pi)r2. (There are two end pieces, so I will be multiplying this area by 2 when I find my total-surface-area formula.) The surface area of the cylinder is the circumference of the circle, multiplied by the height: A = 2(pi)rh.

Side view of the cylindrical tank, showing the radius "r".

An "exploded" view of the tank, showing the three separate surfaces whose areas I need to find.

Then the total surface area of this tank is given by:

2 ×( (pi)r2 ) + 2(pi)rh    (the two ends, plus the cylinder)     = 2( (pi) (42) ) + 2(pi) (4)(30)     = 2( (pi) × 16 ) + 240(pi)     = 32(pi) + 240(pi)     = 272(pi)

Since the original dimensions were given in terms of feet, then my area must be in terms of square feet:

the surface area is 272(pi) square feet.

A piece of 16-gauge copper wire 42 cm long is bent into the shape of a rectangle whose width is twice its length. Find the dimensions of the rectangle.

Do I care that the wire is made of copper, or that the wire is a length of sixteen-gauge? No; all I care is that the length is forty-two units, that the units are centimeters, that the rectangle is twice as long in one direction as the other, and that I'm supposed to find the values of each of these directions. I can ignore the other information.

Since the wire is 42 centimeters long, then the perimeter of the rectangle is 42 centimeters. That is:

2L + 2W = 42

I also know that the width is twice the length, so:

W = 2L

Then:   Copyright © Elizabeth Stapel 2000-2008 All Rights Reserved

2L + 2(2L) = 42    (by substitution for W from the above equation) 2L + 4L = 42 6L = 42 L = 7

Since the width is related to the length by W = 2L, then W = 14, and the rectangle is 7 centimeters long and 14 centimeters wide.

A circular swimming pool with a diameter of 28 feet has a deck of uniform width built around it.  If the area of the deck is 60(pi) square feet, find its width.

I have this situation:

A pool issurrounded by a deck.

The pool has radius 14,and the deck has width "d ".

If the diameter of the pool is 28, then the radius is 14. The area of the pool is then:

(pi)r2 = (pi)(14)2 = 196(pi)

Then the total area of the pool plus the surrounding decking is:

196(pi) + 60(pi) = 256(pi)

Working backwards from the area formula, I can find the radius of the whole pool-plus-deck area:

256(pi) = (pi)r2 256 = r2 16 = r

Since I already know that the pool has a radius of 14 feet, and I now know that the whole area has a radius of 16, then clearly:

the deck is two feet wide.

If one side of a square is doubled in length and the adjacent side is decreased by two centimeters, the area of the resulting rectangle is 96 square centimeters larger than that of the original square. Find the dimensions of the rectangle.

I'm starting from a square with sides of some unknown length. The sides of the rectangle are defined in terms of that unknown length. So I'll pick a variable for the unknown side-length, create expressions for the rectangle's sides, and then work from there.

square's side length: x one side is doubled: 2x next side is decreased by two: x – 2 square's area: x2 rectangle's area: (2x)(x – 2) = 2x2 – 4x new area is 96 more than old area: 2x2 – 4x = x2 + 96

2x2 – 4x = x2 + 96 x2 – 4x – 96 = 0 (x – 12)(x + 8) = 0 x = 12  or  x = –8

I'm supposed to find the dimensions of a rectangle, so can I just erase that one "minus" sign and say that the rectangle is 12 by 8? No! I defined "x" as standing for the side length of the square, not as one of the sides of the rectangle. Looking back at my definitions, I see what "x = 12" (the only reasonable solution for the square) means that the sides of the rectangle have lengths 2(12) and (12) – 2:

The rectangle measures 24 cm by 10 cm

Geometry Word Problems: Triangles (page 3 of 6)

Sections: Introduction, Basic examples, Triangle formulas, Complex examples, The Box Problem & the Goat Problem, Max / Min problems

If the height of a triangle is five inches less than the length of its base, and if the area of the triangle is 52 square inches, find the base and the height.

They have given me a relationship between the height and the base, and have given me the value of the area. So I'll need to use the formula for the area of a triangle with a given base and height, and I'll need to create an expression or equation relating the height and base.

The area of a triangle is given by:

A = ( 1/2 )bh

...where "b" is the base and "h" is the height (or "altitude"). I am given that the height is five less than the base, so the equation for their relationship is:

h = b – 5

Since I am given that the area is 52 square inches, I can then plug the base variable, the height expression, and the area value into the formula for the area of a triangle, and see where this leads:

(1/2)(b)(b – 5) = 52   (by substitution for h from the above equation) b(b – 5) = 104 b2 – 5b = 104 b2 – 5b – 104 = 0 (b – 13)(b + 8) = 0 b = 13  or  b = –8

I can safely ignore the extraneous negative solution. (A solution which is "extraneous", pronounced "ek-STRAY-nee-uss", is a number that is a valid solution to the equation, but is not a relevant value in the context of the word problem. In this case, lengths cannot be negative.) This means that b = 13, so h = b – 5 = 13 – 5 = 8.

The base is 13 inches, and the height is 8 inches.

In the last exercise above, I solved for one value (the length of the base) and then back-solved for the other value (the length of the height). This other value turned out to be the same as the extraneous value, except for the sign change. Warning: Do not assume that you can get both of your answers by arbitrarily changing the sign on the extraneous solution. This does not always work, it is mathematically wrong, it annoys your teacher, and it can get you in trouble further down the line.

Another triangle formula you should remember is the Pythagorean Theorem:

Take a right-angled triangle, and square the lengths of all three sides. If you add up the squares of the two shorter sides, this sum will be the same value as the value of the square of the longest side."

As a formula, the Pythagorean Theorem is often stated in the form "a2 + b2 = c2", where a and b are the lengths of the two legs (the two shorter sides) and c is the length of the hypotenuse (being the longest side, opposite the right angle).

If the sum of the sides of a right triangle is 49 inches and the hypoteneuse is 41 inches, find the two sides.

By "the sides", they mean "the lengths of the two shorter sides". Letting "a" and "b" be the lengths of these sides, the sum is:   Copyright © Elizabeth Stapel 2000-2008 All Rights Reserved

a + b = 49

I can solve this for either one of the variables. I think I'll solve for a in terms of b:

a = 49 – b

This gives me expressions or variables for all three sides of the right triangle: a = 49 – b, b, and c = 41. I'll plug these into the Pythagorean Theorem:

a2 + b2 = c2 (49 – b)2 + b2 = 412     (by substitution) 2401 – 98b + b2 + b2 = 1681 2b2 – 98b + 720 = 0 b2 – 49b + 360 = 0 (b – 9)(b – 40) = 0 b = 9  or  b = 40

In this case, either solution will do. If b = 9, then a = 49 – b = 49 – 9 = 40. Or if b = 40, then a = 49 – b = 49 – 40 = 9. Since the problem didn't specify which of the two legs is longer, it doesn't matter which one I call "a" and which one I call "b". The answer is:

One side is forty inches long, and the other side is nine inches long.

A wood frame for pouring concrete has an interior perimeter of 14 meters. Its length is one meter greater than its width. The frame is to be braced with twelve-gauge steel cross-wires. Assuming an extra half-meter of wire is used at either end of a cross-wire for anchoring, what length of wire should be cut for each brace?

I don't care that the wire is steel; I don't care that they're pouring concrete into a wood frame. All I need is the geometrical information: this is a rectangle with a certain perimeter and a certain relationship between the length and the width. They're asking me, effectively, to find the length of the diameter. And this diameter, together with the length and the width, will form a right triangle. So the perimeter formula for a rectangle may be useful, as may the Pythagorean Theorem.

width: w length: w + 1 perimeter formula: 14 = 2(w + 1) + 2(w)

14 = 2w + 2 + 2w 14 = 4w + 2 12 = 4w 3 = w

Then the length, being one unit larger, is 4, and the Pythagorean Theorem lets me find the length of the diagonal d:

32 + 42 = d2 9 + 16 = 25 = d2 5 = d

Adding a half-meter at either end of the wire, I find that:

each wire should be cut to a length of six meters

Another useful triangle fact is that the measures of any triangle's three angles add up to 180 degrees.

The smallest angle of a triangle is two-thirds the size of the middle angle, and the middle angle is three-sevenths of the largest angle. Find all three angle measures.

The smallest angle is defined in terms of the middle angle, but the middle angle is defined in terms of the largest angle. So it makes most sense to pick a variable for the measure of the largest angle, and then create expressions for the middle and then the smallest angles, using that variable.

I'll let "ß" stand for "beta", the largest angle, or, rather, for the measure of the largest angle. Then the middle angle has a measure of ( 3/7 )ß. The smallest angle is two-thirds of the middle angle, so it has a measure of ( 2/3 )( 3/7 )ß = ( 2/7 )ß. Then my angle-sum formula is:

ß + ( 3/7 )ß + ( 2/7 )ß = 180 7ß + 3ß + 2ß = 1260 12ß = 1260 ß = 105

So the largest angle has a measure of 105 degrees. The middle angle is then:

( 3/7 )(105) = 45

...or 45 degrees, and the smallest angle is:

( 2/3 )(45) = 30

...or 30 degrees.

The angle measures are 30 °, 45 °, and 105 °.

Geometry Word Problems:     Complex Examples (page 4 of 6)

Sections: Introduction, Basic examples, Triangle formulas, Complex examples, The Box Problem & the Goat Problem, Max / Min problems

You work for a fencing company. A customer called this morning, wanting to fence in his 1,320 square-foot garden. He ordered 148 feet of fencing, but you forgot to ask him for the width and length of the garden. Because he wants a nicer grade of fence along the narrow street-facing side of his plot, these dimensions will determine some of the details of the order, so you do need the information. But you don't want the customer to think that you're an idiot, so you need to figure out the length and width from the information the customer has already given you. What are the dimensions?

The perimeter P of this rectangular area with (as-yet unknown) length L and width W is given by:

2L + 2W = 148

The area A is given by:

L×W = 1320

I will divide my "perimeter" equation above by 2, so I am dealing with smaller numbers. This gives me the following system (or "set") of equations:

L + W = 74 L×W = 1320

I can solve either one of these equations for either one of the variables, and then plug this into the other equation. I think I'll solve the addition equation, and plug the result into the multiplication equation:

L = 74 – W   (solving the first equation for L) (74 – W) × W = 1320    (substituting into the second equation) 74W – W 2 = 1320 0 = W 2 – 74W + 1320 0 = (W – 30)(W – 44) W = 30  or  W = 44

Once again, I've come up with two valid solutions. If W = 30, then L = 74 – W = 74 – 30 = 44.  If W = 44, then L = 74 – W = 74 – 44 = 30. The important point is that the shorter side (whether I refer to it as the "width" or the "length") is across the front of the lot.

The garden is 44 feet by 30 feet, with the 30-ft length along the front.

Note that we cannot say which of the dimensions is the length or the width, since no information was provided regarding which was longer. So "this by that" is as accurate an answer as we can give.

Three times the width of a certain rectangle exceeds twice its length by three inches, and four times its length is twelve more than its perimeter. Find the dimensions of the rectangle.

The first statement, "three times the width exceeds twice its length by three inches", compares the length L and the width W. I'll start by doing things orderly, with clear and complete labelling:

the width: W three times the width: 3W twice its length: 2L exceeds by three inches, meaning "is three inches greater than": + 3 equation: 3W = 2L + 3

The second statement, "four times its length is twelve more than its perimeter", compares the length L and the perimeter P. I will again be complete with my labelling:

four times its length: 4L perimeter: P = 2L + 2W     (this is just the perimeter formula for rectangles) twelve more than: + 12 equation: 4L = P + 12, or 4L = (2L + 2W) + 12  (by substitution)

So now I have my two equations:   Copyright © Elizabeth Stapel 2000-2008 All Rights Reserved

3W = 2L + 3 4L = 2L + 2W + 12

There are various ways of solving this; the way I do it (below) just happens to be what I thought of first. I'll take the first equation and solve for W:

3W = 2L + 3 W = ( 2/3 )L + 1

Now I'll simplify the second equation, and then plug in this above expression for W:

4L = 2L + 2W + 12 2L = 2W + 12 2L = 2[ ( 2/3 )L + 1 ] + 12    (by substitution from above) 2L = ( 4/3 )L + 2 + 12 2L = ( 4/3 )L + 14 2L – ( 4/3 )L = 14 ( 6/3 )L – ( 4/3 )L = 14 ( 2/3 )L = 14 L = (14)×( 3/2 ) = 21

Then:

W = ( 2/3 )L + 1     = ( 2/3 )×(21) + 1     = 14 + 1 = 15

The question didn't ask me to "Find the values of the variables L and W". It asked me to "Find the dimensions of the rectangle," so the actual answer is:

The length is 21 inches and the width is 15 inches.

Geometry Word Problems:     The Box Problem & The Goat Problem (page 5 of 6)

Sections: Introduction, Basic examples, Triangle formulas, Complex examples, The Box Problem & the Goat Problem, Max / Min problems

The following are a couple of "classic" types of exercises. Almost all students eventuall see one or both of these, in some form.

You need to make a pizza box. You know that the box needs to be two inches deep, it needs to be a square, and the web site you found said that the box needs to have a volume of 512 cubic inches. After cursing the occasional near-uselessness of the information you find on the Internet, you start calculating the dimensions you will need.

You have a large piece of cardboard, but you don't have enough cardboard to make a mistake and try again, so you'll have to get it right the first time. You will be forming the box by cutting out a large square, and then cutting out the two-inch squares from the corners that will allow you to fold up the edges to make a two-inch-deep box. What should be the dimensions of the large square? (Ignore the top of the box:  you'll just make another open box, slightly larger, turn it upside down, and slip it over the first box to make the "top".)

Visually, this is what I'm doing:

This square stands for my initial piece of cardboard, out of which I'll be cutting

corner squares and folding up the sides.

The square's dimensions are x inches by x inches, but I don't know the value

of x yet.

Now I have cut the corners out, leaving two-

inch flaps on all four sides. I will be folding the sides up along those red lines.

Now I have folded up the flaps to make a two-inch

deep box. (I will be taping the

corners to hold them together.)

I will use "x" to stand for both the length and the width of the original square of cardboard in the computations that follow. The value of x is what I'm looking for.

Looking at the picture, I can see that the width of the bottom of my box will be x – 2 – 2; that is, the final width of the box will be the sheet's original x inches, minus two inches on either side because of the portions that I'll be losing to the flaps I'll be folding up. Then the width of the bottom of my box is going to be x – 4. By the same reasoning, the length will also be x – 4.

Since the depth of my box is 2 (that's the whole point of the two-inch-by-two-inch squares that I'm cutting out of the corners), I can write the formula for the volume of this box as:

volume = (length)(width)(depth) = 512 cubic inches (x – 4)(x – 4)(2) = 512 2(x2 – 8x + 16) = 512 x2 – 8x + 16 = 256 x2 – 8x – 240 = 0 (x – 20)(x + 12) = 0 x = 20  or  x = –12

I can ignore the extraneous negative result. Then x = 20 inches, and the answer is:

The large piece of cardboard should be twenty inches square.

Warning: "Twenty inches square" is not the same as "twenty square inches". "Twenty inches square" means "twenty inches on a side", for a total area of four hundred square inches. "Twenty square inches" is just that: twenty square inches.     Copyright © Elizabeth Stapel 2000-2008 All Rights Reserved

A goat is tied to the corner of a 5-by-4-meter shed by a 8-meter piece of rope. Rounded to the nearest square meter, what is the area grazed by the goat?

What formula could they be expecting me to use for this? How on earth would I answer this? When in doubt, it can be helpful to draw a picture, so let's see what I can find:

The goat is tied at the upper right-hand corner with the dot. Now what? Well, on the nearest two sides, the goat can go this far:

This is three-quarters of a circle with radius 8 (the full length of the rope). The goat can walk around the far corners as far as the rope will allow. Along the top side, five meters of rope will be stretched along the side, leaving another three meters "in play"; along the right-hand side, four meters will have been used, leaving another four meters:

These new areas are both quarter-circles, one with radius 3 and the other with radius 4. And there will be no way for the goat to reach the opposite corner or "overlap" in grazing, because he's out of rope. So these three partial circles mark off the total grazing area.

I know the formula for the area of a circle. To find, say, the area of 3/4 of a circle, I'd just multiply the total area by 3/4. So the total area is:

(3/4)(pi)(82) + (1/4)(pi)(32) + (1/4)(pi)(42)      = 48pi + 2.25pi + 4pi = 54.25pi

I'm supposed to give my answer in terms of the nearest whole unit, so I'll plug the above into my calculator, and round:

The goat can graze about 170 square meters of grass.

Note how drawing the picture allowed me to "see" what I needed to do, immediately simplying my work. If you have "no idea" what to do, try drawing a picture. (There is an extension of this exercise, which has the goat's areas overlapping at that opposite corner. The exact solution involves trigonometry or calculus.)

Geometry Word Problems:     Maximizing and Minimizing (page 6 of 6)

Sections: Introduction, Basic examples, Triangle formulas, Complex examples, The Box Problem & the Goat Problem, Max / Min problems

A special case of quadratic-based geometry word problems involves having to maximize or minimize some dimension. This will involve finding the vertex of the quadratic formula you come up with, since the vertex will be the maximum or minimum of the graph.

Find the largest possible rectangular area you can enclose, assuming you have 128 meters of fencing. What is the (geometric) significance of the dimensions of this largest possible enclosure?

I'll let the length be L and the width be W. I have 128 meters of fencing, so the perimter equation is: 

2L + 2W = 128

Dividing by 2 to make things simpler, I get:

L + W = 64

Previously, they would have given me the area and I would have had to find the length and width. This time, they told to find the area; in particular, to find the largest area, given this perimeter. How do I do that? Let's look at the area equation:

A = L × W

I can substitute for either one of these variables by solving the perimeter equation:

L + W = 64 L = 64 – W      (solving for L)

Then:

A = (64 – W) × W     (substituting into the area equation)      = 64W – W 2

In other words, my area equation is a quadratic, and I'm supposed to find the maximum. So all I really need to do is find the vertex. Since the above area equation is a negative quadratic, then it graphs as an upside-down parabola, so the vertex is the maximum.

There are a couple different ways of finding the vertex. I'll take the easy way. The equation of the quadratic, in y = ax2 + bx + c format, is:

A = –W 2 + 64W

The vertex of a parabola is the point (h, k), where h = –b/2a .  In this case:

h = –(64)/(2×(–1)) = 32

To find the "k" part of the vertex, all I do is plug 32 in for W:

k = –(32)2 + 64(32) = 1024

My points from this equation are (W, A) — that is, I plug in a width and figure out the area — so the "h" is the maximizing width and the "k" is the maximum area. So the answer is:

The largest possible area is 1024 square meters

...and I know that this maximum occurs when the width is 32 meters. Now I also need to find the length, because the original question asked about the "significance" of the dimensions. Since W = 32, then:     Copyright © Elizabeth Stapel 2000-2008 All Rights Reserved

L = 64 – W = 64 – 32 = 32

Then the length and width are the same: 32 meters. What do you call a rectangle that is as wide as it is long? A square. So the second part of the answer is:

The largest possible rectangular area is in the shape of a square.

Educators have started noticing that students have figured out the solution to the above exercise, just as a rule: "The rectangle with the largest area for a given perimeter will be a square" and, vice versa, "The rectangle with the shortest perimeter for a given area will be a square". So they've come up with new forms of the exercise; fortunately, the reasoning and general process is exactly the same.

The riding stables has just received an unexpected rush of registrations for the next horse show, and quickly needs to create some additional paddock space. There is sufficient funding to rent 1200 feet of temporary chain-link fencing. The plan is to form two paddocks with one shared fence running down the middle. What is the maximum area that the stables can obtain, and what are the dimensions of each of the two paddocks?

To help me "see" what I'm doing, I first draw a picture:

The total area A for the two paddocks will obviously be A = Lw. The total length of fencing gives me the total "perimeter" (in quotes, because I'm also including that shared line down the middle, so this isn't the "regular" perimeter). Then:

P = 1200 = 2L + 3w

I can't do much with this, but what if I solve this for one of the variables, and then plug that into the "area" formula? Let's see...

1200 = 2L + 3w 1200 – 3w = 2L + 3w – 3w 1200 – 3w = 2L (1200 – 3w) / 2 = (2L) / 2 600 – (3/2)w = L

Now I'll plug this into the "area" formula:

A = Lw = [600 – (3/2)w](w) = 600w – (3/2)w2    = (–3/2)w2 + 600w

This is a negative quadratic, just like the previous exercise, and I'll find the maximum area in the same way: by finding the vertex.

h = –b / (2a) = –(600) / [2(–3/2)] = –600 / –3 = 200

So I get the maximum area when the input (the width, in this case) has a value of 200. Reviewing my picture and equations, I see that the width should be 200 feet, the overall length should be 300 feet, and each paddock should then be 150 feet long.

The maximum added area will be 60,000 square feet (sq ft), 30,000 sq ft for each paddock. Each paddock should measure 200 ft by 150 ft, and the paddocks should share a 200-ft long side.

In particular, note that the maximal area above is not a square! Other ways of skewing the solutions away from squares, circles, or spheres is to include cost considerations, such as the material for the base of an open-topped box costing more (because it needs to be stronger) than the material for the sides. Don't just assume that the "neatest" answer will be correct. Do the math.