Find the value of x for which the area is a maximum:
A = 800 + 20x - 1/2x^2
find x for when A=0
it is a quadratic, you get two values of x. Because the maximum is half way between these two minimums, (it is a parabola), then the max is at the sum of the roots divided by 2.
put that average x into the formula, and you have the max value.
do I use the quadratic formula to find the two x values?
yes, or complete the square
-(1/2) x^2 + 20 x = A - 800
x^2 - 40 x = -2 A + 1600
x^2-40x+20^2 = - 2A + 2000 = -2(A-1000)
(x - 20)^2 = -2 (A-1000)
vertex at x = 20, A = 1000
To find the value of x for which the area A is a maximum, we need to differentiate the equation with respect to x and set the derivative equal to zero. This will give us a critical point, which could correspond to the maximum area.
Let's differentiate the equation with respect to x:
dA/dx = d(800 + 20x - 1/2x^2)/dx
To differentiate, we need to apply the power rule and constant multiple rule:
dA/dx = 0 + 20 - (1/2)(2x)
Simplifying further:
dA/dx = 20 - x
Now, set the derivative equal to zero and solve for x:
20 - x = 0
x = 20
Therefore, the value of x for which the area is a maximum is x = 20.