Posted by **Lucas** on Thursday, October 24, 2013 at 12:32am.

A company's weekly profit, in riyals, is modeled by the function P(u)=-0.032u^2+46u-3000. where u is the number of units sold each week.

a) the maximum weekly profit. Answer: 13531.25 riyals

b) the loss for week's holiday period, where no units are sold .Answer:3000 riyals

c) the number of units sold each at break-even point for the company. Answer: 69 or 1369 units

Thank you so, so much for everything. Really thanks a lot for explanation of this problems. Answers are from the book.

- math -
**Steve**, Thursday, October 24, 2013 at 4:23am
P(u) is a parabola, with vertex at u=718.75

Plug in that value and verify P.

(b) easy to read off, if u=0

(c) breakeven when P=0. Just solve the equation for u. When rounded, will yield the given answers.

- math -
**Jai**, Thursday, October 24, 2013 at 4:29am
(a)

To get the maximum profit, we differentiate the given function with respect to u. And then equate everything to zero (because at maximum, the slope is zero):

P = -0.032u^2 + 46u - 3000

dP/du = -0.064u + 46

0 = -0.064u + 46

0.064u = 46

u = 718.75 (this is the number of units that will yield the max profit)

Substituting to the given function,

P = -0.032u^2 + 46u - 3000

P = -0.032(718.75)^2 + 46(718.75) - 3000

P = 13531.25 riyals

(b)

This is easier. We only let u = 0 (because this is the lowest number of units that will yield the lowest profit, when the company sold nothing):

P = -0.032u^2 + 46u - 3000

P = -0.032(0)^2 + 46(0) - 3000

P = -3000 riyals

(c)

Solve for the value of u at P = 0. This value of u will yield no profit nor loss:

P = -0.032u^2 + 46u - 3000

0 = -0.032u^2 + 46u - 3000,

Solving,

u = 1369.02 and u = 68.48

Hope this helps :)

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