• Jul 27th 2013, 07:23 PM
Perezjay
Suppose that the demand x(in units) for a product is x=10,000-100p, where pd dollars is the market price per unit. The consumer expenditure for the product is E=px=10,000p-100p^2

For what market price will expenditure be greatest?
• Jul 29th 2013, 11:28 AM
ebaines
Are you familiar with finding the min or max of a function by setting its derivative equal to zero? Here you have E = 10000p-100p^2, so dE/dp = 10000-500p. Set this equal to zero, solve for p, and check that the result is a max and not a min by seeing if d^2E/dp^2 is positive or negative.

If you are not famiiar with calculus you can try plotting the graph for E=10000p-100p^2 and see if you can estimate what value of p makes E the largest.
• Jul 29th 2013, 12:27 PM
HallsofIvy
This is a quadratic function that does not require the "derivative" (or even knowing what a "derivative" [b]is[b]) to find max or min. Instead "complete the square". A "perfect square" is of the form $(x- a)^2= x^2- 2ax+ a^2$. Compare that to $10000p- 100p^2= -100(p^2- 100p)$. Since "p" corresponds to "a", "-100p" corresponds to -2ax, thus a corresponds to 50, and then $a^2= 2500$. To make $p^2- 100p$ a perfect square we need to add 2500. And, of course, to keep the actual value the same, we need to subtract 2500 as well.

That is, $-100(p^2- 100p)= -100(p^2- 100p+ 2500- 2500)= -100(p^2- 100p+ 2500)+ 250000= -100(p- 10)^2+ 25000$.

Recalling that a square is never negative, it follows that $-100(p- 10)^2$ is never positive. So what is the maximum value for consumer expenditure and for what p does it have that maximum?
• Jul 29th 2013, 02:22 PM
ebaines
$-100(p^2- 100p) .... = -100(p- {\color{red}5}0)^2+ 25000{\color{red}0}$.
Recalling that a square is never negative, it follows that $-100(p- {\color{red}5}0)^2$ is never positive.... "