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?

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- Jul 27th 2013, 06:23 PMPerezjayPlease help!
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, 10:28 AMebainesRe: Please help!
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, 11:27 AMHallsofIvyRe: Please help!
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 $\displaystyle (x- a)^2= x^2- 2ax+ a^2$. Compare that to $\displaystyle 10000p- 100p^2= -100(p^2- 100p)$. Since "p" corresponds to "a", "-100p" corresponds to -2ax, thus a corresponds to 50, and then $\displaystyle a^2= 2500$. To make $\displaystyle p^2- 100p$ 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, $\displaystyle -100(p^2- 100p)= -100(p^2- 100p+ 2500- 2500)= -100(p^2- 100p+ 2500)+ 250000= -100(p- 10)^2+ 25000$.

Recalling that a square is never negative, it follows that $\displaystyle -100(p- 10)^2$ is never positive. So what is the maximum value for consumer expenditure and for what p does it have that maximum? - Jul 29th 2013, 01:22 PMebainesRe: Please help!
Interesting technique - thanks for that. But a couple of typos crept into the equation. I've taken the liberty of making the corrections below in red:

"That is,

$\displaystyle -100(p^2- 100p) .... = -100(p- {\color{red}5}0)^2+ 25000{\color{red}0}$.

Recalling that a square is never negative, it follows that $\displaystyle -100(p- {\color{red}5}0)^2$ is never positive.... "