Question

A(x)=x(10-x)
When is A(x) at its maximum? Explain
Or show how you know.

Answer 1

Answer:

A(x) is at its maximum at x = 5, or (5, 25)

Step-by-step explanation:

Putting A(x)=x(10-x) back into its standard form gives A(x) = -x^2 + 10x. The equation to find the x value in vertices in these types of functions is -b / 2a, the letters pertaining to the coefficients. In this case, this graph would have a maximum at -(10)/2(-1), or -10/-2 = 5 --> Thus, when x=5, the downwards facing graph is at its maximum

(Might not be important but another method to solve:)
Take the derivative of the function A(x) = x(10-x) and set it equal to 0 to find the x value of all vertices. In this case, A'(x) = -2x + 10,
-2x + 10 = 0
-2x = -10
x = 5
A(x) is at its maximum at x=5 because it is when A'(x), the derivative, is equal to 0 and thus has a flat slope (or vertex)

Answer 2

Answer:

(5, 25)

Step-by-step explanation:

A(x) is a parabola meaning the max point would be called the vertex

Expand A(x)

$x(10 - x) = 10x - x^{2}\\x^{2} -10x$

The vertex point is (h, k)

To find h:

$h =- \frac{-b}{2a} \\b = -10\\a = 1\\h = - \frac{-10}{2(1)} \\h = - \frac{-10}{2} \\h = 5$a and b are from the equation $ax^{2} + bx$

To find k, plug in h values:

$k = \\f(5) = 5^{2} - 10(5)\\f(5) = 25$

vertex = (h, k) = (5, 25)

graph attached as well for a visual