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+46u3000. 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 breakeven 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 :)