math
posted by Lucas on .
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.

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. 
(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 :)