Consider the function $f(x)=9−x^2$ for $f(x)≥0$ only.
Let $T(x)$ equal the area of the shaded isosceles trapezoid that has two vertices on the x-axis and two vertices on the graph of $f$, as shown.
What is the maximum value of $T(x)$?
Question was taken from Khan Academy
Steps I took:
I know that the area of a trapezoid is $\frac { a+b }{ 2 } \cdot h$
In this example, it seems that the height $h$, is $f(x)$ and the width of the top parallel side is $2x$
So I end up with $T(x)=\frac { 2x+b }{ 2 } (9-x^ 2)$
Now I know that I must find the equation of the area of the trapezoid, then find it's derivative, set it to $0$ to find the max and plug that max back into the original area equation. I just don't know how to get past the current step I am at.
Best Answer
You have most of the steps. The area can be defined as:
$$A = \frac{2x + b}{2}(9-x^2)$$
Now we must find $b$. Note that we are looking for the maximum area, so choosing the x-intercepts of the functions, as you have indicated in your picture above, will give us the maximum possible area.
Since the zeros of the function are $x = -3, 3$, $b = 3 + 3 = 6$
Now your area is:
$$A = \frac{2x + 6}{2}(9-x^2)$$
Taking the derivative, we get:
$$A' = -3x^2 -6x +9 = 0$$
Solving for $x$ we get:
$$-3(x+3)(x-1) = 0$$
Therefore, $x = -3, 1$
Testing the points, we find that $x=1$ yields the maximum value:
$$A = \frac{2 + 6}{2}(9-1) = 32$$