Are the derivatives of each accurate?

10 180/p = -180/p^2

0.1/x^2 = -0.2/x^3

Dont know what the 10 in front of the 180 is for...

but the 2 answers are correct

ignoring the 10 in front of the first question (maybe it is the question number?), it looks okay.

Answer mine:

a company is producing parts to sell to an auto parts stoe. The store buys in batches of 100, and pays the company $800 for each batch of 100 sold. The cost for the company to make x parts is modeled by the function c(x) = x^(3/2)/12 + 5x -10. To the neaest whole dollar, what is the maximum possible profit for the company.

Answer is 585, but I'm getting 586. Multiple choice question and both answers are shown.

Thanks guys. And the 10 was supposed to be "10 plus 180" but the plus symbol disappeared. I know the constant goes to 0 so I could have left it out.

buckshot44 -

lol we had to do problems like that then our teacher got sick and we had a sub for 4 months. i got a 100 in calculus :)

kungfugrip - Answer mine:

a company is producing parts to sell to an auto parts stoe. The store buys in batches of 100, and pays the company $800 for each batch of 100 sold. The cost for the company to make x parts is modeled by the function c(x) = x^(3/2)/12 + 5x -10. To the neaest whole dollar, what is the maximum possible profit for the company.

Answer is 585, but I'm getting 586. Multiple choice question and both answers are shown.

It's relatively easy to figure out in Excel, but harder using calculus. Set up a profit as a function of y, where y is a batch of 100 units. Profit = 800*y-[(100y)^(3/2)/12+500y-10]

Differentiate, set equal to zero, solve for y, and you get y = 5.76

Since Y is discreet, check the profit of 5 and 6 batches. 6 Batches = Profit of 585.26, 5 batches = Profit of 578.31

So the answer is 585

tdunning,

Thanks! I set it up correctly (didnt change x to 100y) but didnt think about the meaning of the numbers and just plugged in 576 instead of "whole" batches of 500 or 600.

Thanks again!