# Calculus “Word Problem”

Here is the problem:

A rectangle has its base on the $$x$$-axis and its upper two vertices on the parabola $$y=12-x^2$$ What is the largest area the rectangle can have, and what are its dimensions?

Well, I don't really know where to start. My initial idea was to find inflection points because I figured that is where the vertices would be, but there are no inflection points because it is a parabola.

Then I though about finding where the derivative and the parabola cross, found it, but I don't know how that will help me.

I really don't know where to start.

Any help is appreciated, thanks.

• You should be able to express the area of the inscribed rectangle in terms of $x$. Differentiate that, and find the values of $x$ that make it zero, and check that your results do correspond to a maximum. – J. M. is a poor mathematician Dec 21 '11 at 16:57
• Okay, so: xy=A, X(12-x^2)=A, -3x^2 + 12 = 0, x = 2, y = 8, Area = 16. – user21589 Dec 21 '11 at 17:06
• @user21589: Almost there! Area is $2x(12-x^2)$, base goes from $-x$ to $x$. If you have doubts, picture will take care of them. You found the area of the first quadrant part of the best rectangle. – André Nicolas Dec 21 '11 at 17:17

Have a look at following picture. Does this help? Can you get (using this picture) area of the rectangle?

You should get $A(x)=2x(12-x^2)$ as mentioned in André's comment.

After getting this expression for $A(x)$, can you continue with the solution? What will be $A'(x)$ and where is it equal to zero?

You have $A(x)=24x-2x^3$, which means $$A'(x)=24-6x^2.$$ Solving $A'(x)=0$ gives $x=2$ and $A(x)=2\cdot2\cdot(12-2^2)=2\cdot2\cdot8=32$ as you correctly stated in the comment.

(There is also solution $x=-2$ but that one is not interesting for us - we are looking for $x\ge 0$ since we denote by $x$ the point to the right from $0$.)

• Whether I use 2x or x, I still get x=2 in the end, and y = 8, thus area = 16. Did I do something wrong? Assuming above equation, next step is 24x-2x^3 = A, -6x^2 + 24 = 0, x=2, y=8, area=16. – user21589 Dec 21 '11 at 17:30
• The only think to do is multiply the area by 2, to get the whole rectangle. Otherwise your solution is fine. (It should be also clear that you should get the same $x$ when you maximize the area of rectangle or half of this area.) – Martin Sleziak Dec 21 '11 at 17:35
• Ah, I see exactly what you mean now. Thanks. – user21589 Dec 21 '11 at 17:38

The function $y=12-x^2$ is symmetrical to the $y$-axis, so in order for the upper corners of the rectangle to be on the curve, they must be at x-values that are opposites.

The length of the bottom of the rectangle is $2x$.

The height is $12-x^2$.

So the area of the rectangle at a value $x$ is:

$$A(x) = 2x(12-x^2) = 24x-2x^3.$$

$$A'(x) = 24-6x^2.$$

We set this equal to zero to determine the $x$-value which makes $A(x)$ maximum.

$$24-6x^2=0 \Rightarrow 4=x^2\Rightarrow x=\pm 2.$$

We take $x=2$.

So the area is maximized when the bottom corners are at $(2,0)$ and $(-2,0)$.