Otter Pop Frenzy

By Karly Kelso

Costco is revamping their computer programs, but their

system is currently shut down.

They need to negotiate a new contract with the makers of Otter Pops immediately, and they only have a few random graphs with

equations from a previous presentation to look at.

If Otter Pops were sold year round, the amount of people buying them

can be shown as

y(t)=-2.413x³+24.715x²+86.307x-227.867where t is the month and t=1 is January

How many Otter Pops could potentially be sold?

ANSWER

The way to figure out this answer is to look at the area

under the curve or the integral of the equation.

If you take the integral of y(t)=-2.413x³+24.715x²+86.307x-227.867

You get:-.603x4+8.238x³+43.154x²-227.867x│ from 1 to 12

plugging in Y(12)-Y(1) gives the answer of approximately 5384 boxes of Otter Pops

You can also plug this equation into your calculator and graph it and then hit 2nd Calc and 7 to calculate the

integral from 1 to 12.

If sold at $9.49 per box (which happens to be the highest price they can be sold at), how much

money would be made if the total potential sold could be

reached?

ANSWER

This problem is a little more simple. You simply multiply the number you reached in

part A by 9.49 to get the answer $51,094.16

HOWEVER, the Otter Pop company will sell Otter Pops at a

lower price when the quantity bought is higher, but the boxes can

only be purchased during the month that Costco is planning on

selling them…

The price of each Otter Pop when bought in a bulk is modeled by the equation

y(b)=21.525-1.978lnbwhere b is the total number

purchased and y(b) is the price per Otter Pop box.

If 10% profit must be made per box, when can Costco afford to purchase these Otter Pops from

the company?

ANSWER

First you must calculate the highest price that the Otter

Pops can be bought at to earn a profit.

This amount will be equal to 9.49/1.1 because the general equation is 110% X the price

bought=the price sold at.So the price bought=the price

sold at/110%(or 1.1)This equals $8.63

Now the next trick to solving the equation is to set $8.63 equal to

the equation y(b)=21.525-1.978lnb

8.63=21.525-1.978lnb-12.895=-1.978lnb

6.519=lnbe6.519=b678=b

This value of b says that Otter Pops must be purchased in

orders larger than 678 in order to earn enough profit. Now you must set THIS number equal toy(t)=-2.413x³+24.715x²+86.307x-227.867

to find out which months are profitable

It is profitable when the equation is greater than 678 so you set this equation equal to

678 and solve for x.

However, an easier way to solve the problem is to graph

y(t)=-2.413x³+24.715x²+86.307x-227.867 and y=678

and use your calculator to find the intersection points of these

two graphs

When you do this, you find that the intersection points are

AROUND 6 and 10, meaning that Otter Pops should be sold

between June and Octoberin order to make a high enough

profit

The Otter Pop company also wanted to know for which month they should expect the largest

order from Costco

For this, you must find the derivative of the original

equation

Original equation y(t)=-2.413x³+24.715x²+86.307x-227.867

Derivative of equationy’(t)=-7.239x²+49.43x+86.307

The month for the largest amount of Otter Pops sold is a

maximum. To find the maximum, check the endpoints,

where the derivative equals zero, and where the derivative

does not exist.

To find where the derivative equals zero, you graph the

equation and hit 2nd calc and 2. This produces an answer of 8.3.

Also, the derivative exists everywhere.

So now you must check where x=1,8.3, and 12

x=1, y=-119.3x=8.3, y=811.38x=12, y=197.11

Therefore, 8.3 is a maximum so August is the month where the most number of Otter Pops are

sold.