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 $T(x)$?

enter image description here

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.


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:

$$A = \frac{2 + 6}{2}(9-1) = 32$$

  • $\begingroup$ beautiful explanation! Thank you! $\endgroup$ – Cherry_Developer Feb 9 '15 at 11:50
  • $\begingroup$ I didn't realize I could find the value of $b$ by finding where they intercept the x-axis $\endgroup$ – Cherry_Developer Feb 9 '15 at 11:51

You are now at


You can then find the derivative of it and set it equal to zero.

  • $\begingroup$ how did you get $(x+3)$ $\endgroup$ – Cherry_Developer Feb 9 '15 at 11:40
  • $\begingroup$ I do not think that $b =2$ in this case, rather it is $3$ instead $\endgroup$ – Varun Iyer Feb 9 '15 at 11:41
  • $\begingroup$ Since your $b$ is actually $6$. It is the distance between the two $x$ intercepts. $\endgroup$ – KittyL Feb 9 '15 at 11:41
  • $\begingroup$ @KittyL my mistake, you are correct. $\endgroup$ – Varun Iyer Feb 9 '15 at 11:42

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