The largest possible area of the rectangle on a parabola

Question

A rectangle has its base on the x-axis and its upper two vertices on the parabola $y = 49 - x^2.$ What is the largest possible area (in square units) of the rectangle?

I do not know how to write the area formula though, I am guessing that it is $x \times y = x \times (49 - x^2)$ am I correct? and then I have to differentiate and equate the differentiation to zero after that.

My question is my formula for the area correct? or it needs to be multiplied by 2? if so why?

Answer 1

Always draw a picture.

If you assume the rectangle has horizontal sidelength $2x$, then this is the picture:

The area of the rectangle is $f(x) = 2x \cdot (49-x^2)$. This function is maximal when $f^\prime(x) = 98 - 6x^2= 0$, i.e. when $x=\frac{7}{\sqrt{3}}$.

Thus the maximal area is $$f\left ( \frac 7 {\sqrt{3}}\right) = \frac{14}{\sqrt{3}}\left( 49-\frac {49}3\right)=\frac{1372}{3\sqrt 3}$$

Answer 2

$A = 2 x (49 - x^2)$

Find $x$ s.t. $\frac{d A}{d x} = 0$. (Answer: $x = 7/\sqrt{3}$)...

Can you finish from there?

Answer 3

Notice that in order to be a rectangle, its upper vertices must be $(x, 49-x²)$ and $(-x,49-x²)$, because a point $A$ and a point $B$ in a parabola will only form a horizontal straight line if $B$ is the vertical reflection of $A$. Its lower vertices will have to be $(x,0)$ and $(-x, 0)$, because they are the perpendicular projection of the upper vertices on the $x$-axis. So, the length of the basis is $x-(-x)=2x$, therefore you must multiply by two. With that in mind, just apply the same algorithm you would before: multiply $2x$ and $y$, differentiate it, and then equal it to 0.