p(x)=800−0.5x find the maximum

Answer 1

it is a linear function with no max or min....

quadratics however have a max or min...

so IF p(x) = 800 - 0.5x^2 then

p(x) = -0.5x^2 + 800

A= -0.5 , B=0, C=800

-b/(2a) = 0 / (2 * -0.5) = 0

So the extrema occurs at x=0...

p(0) = -800

Note that the extrema also occur at the AVERAGE of the solutions...

the solutions are: -0.5(x^2-1600)=0

x^2 - 1600 = 0

x^2 = 1600

x = 40 and x=-40

the average of which is zer0.

the min point is (0,-800)