Question

What is the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve y = 27 − x^2?

324
108
18
3

Answer 1

Let's try to tease out a function for the area of our hypothetical rectangle: 
We know that the area of a rectangle is Base x Height, and the base will be the length of the x-axis portion of the rectangle. Looking at a graph of y=27 - x^2 will help with intuition on this. 

The length of the base will be 2x, since it will be the distance from the (0,0) axis in the positive direction and in the negative direction. 
So our rectangle will have an area of 2x, multiplied by the height. 
What is the height? The height will be our y value. 
Therefore, 
A = 2x * y, where x is x-value of the positive vertex. 

We already know what y is as a function of x:
y= 27 - x^2
That means our equation for the area of the rectangle is:
A = 2x (27 - x^2)           Distribute the terms....
A = 54x - 2x^3      
This is essentially a calculus optimization problem. We want to find the Maximum for A, so let's find where the derivative of A is equal to zero. 
First, we find the derivative: 
A = 54x - 2x^3
A' = 54 - 6x^2 
Set A' equal to zero to find the maximum value for A
0 = 54 - 6x^2
6x^2 = 54
x^2 = 9
x = 3
We got our x-value! Now let's find the y value at that point:
y= 27 - x^2
y = 27 - 9
y = 18

The height our rectangle will be 18, and our base will be 2*x = 2*3 = 6
Area = base x height = 18 * 6 = 108
The answer is B) 108.

Answer 2

to the risk of being redundant

check the picture below.

so, notice, if we pick a point "a" on the x-axis, then the width of such rectangle is a+a, or 2a, and the length is 27-a²,

thus

$\bf A(a)=2a(27-a^2)\implies A(a)=54a-2a^3 \\\\\\ \cfrac{dA}{da}=54-6a^2\impliedby \textit{let's check for critical points} \\\\\\ 0=54-6a^2\implies 6a^2=54\implies a^2=\cfrac{54}{6}\implies a^2=9 \\\\\\ a=\pm\sqrt{9}\implies a=\pm 3$

if you run a first-derivative test on the region before -3, between -3 and 3, and after three, say for example do a check on f'(-4), and f'(0) and f'(4), you'll see the -3 is a minumum, whilst the +3 is a maximum.

therefore A(a) has an absolute maximum at 3, thus the largest area is then A(3), and surely you know what that is.