A rectangle has two of its vertices as the $x$-axis. The other two vertices are on the parabola whose equation is $y=18-x^2$.
What are the dimensions of the rectangle if its area is to be a maximum?
[Math] How to find the dimensions of a rectangle if its area is to be a maximum
calculus
Related Solutions
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 $0$.)
We have
$$P = 2(l+w) \implies 100 = 2(l+w) \iff \color{blue}{50 = l+w} \tag{1}$$
$$\color{green}{A = lw} \tag{2}$$
Rewriting $(1)$ in terms of $w$ (you could do so in terms of $l$), you’d get
$$w = 50-l$$
Plugging $w = 50-l$ in $(2)$, you have
$$A = l(50-l)$$
$$\color{purple}{A = 50l-l^2} \tag{3}$$
You’ve gotten it correctly until here. Treat $A$ as a function. The resulting quadratic is concave down, so the vertex is a maximum.
$$l_{vertex} = \frac{-b}{2a} = \frac{-50}{2(-1)} = 25$$
Plugging $l = 25$ in $(1)$, it becomes clear that $w = 25$. Hence, the maximum area occurs when there is a square, so
$$\boxed{A = s^2 = 25^2 = 625}$$
Using only quadratics is faster than using optimization, but that is of course correct as well:
$$\frac{dA}{dl} = 50-2l$$
$$\frac{dA}{dl} = 0 \implies 50-2l = 0 \implies l = 25$$
from which the same answer is obtained.
In response to your second question, you can’t find a minimum area, because it simply doesn’t exist. (Unless if you set $w = 0$ in which case the area becomes $0$.) A concave down parabola has a maximum, not a minimum. A minimum point is found when the parabola is concave up, which isn’t the case here.
Best Answer
The equation for the area of the rectangle (depending on the $x$-coordinate of the vertex on the right side of the $y$-axis) will be $A(x)=2x(18-x^2)$. Do you know how to determine which $x$ maximizes this? Note that we will only be considering $0<x<\sqrt{18}$.
Edit: The first thing we probably want to do for a problem like this is draw a picture. A general rectangle as described will look something like this:
Note that the height of the rectangle is $y$, and the width is $x-(-x)=2x$, so the area is $2xy$. We also know $y=18-x^2$, since $(x,y)$ lies on that parabola, so that gives us the expression I have above for area, and it's entirely in terms of $x$. We can also rewrite it as $A(x)=36x-2x^3$. Taking the derivative, we have $A'(x)=36-6x^2$. We need to find the positive solution to $A'(x)=0$, which will give us the maximum area. If $x_0$ is that value of $x$, then the dimensions will then be $2x_0$ by $18-x_0^2$.