Question

WILL BECOME A FAN AND GIVE A MEDAL
What is the largest rectangle with lower base on the x-axis and upper vertices on the curve y=12-x^2?
(A) 8
(B) 12
(C) 16
(D) 32
(E) 48

Could you please show your work as well as select an answer, thank you so much!

Answer 1

First, let's state what we know. We know the general area of a rectangle and the curve we are given, so let's put them down:

y = 12 - x^2
Area = xy

Now, substitute in what we're given from the curve into the Area formula:

Area = x(12 - x^2) = -x^3 + 12x

To find the max area, we should differentiate and set it equal to zero to solve for our critical value(s).

Area ' = -3x^2 + 12
Area ' = 0 = -3x^2 + 12
-12 = -3x^2
4 = x^2
x = +/- 2

Now, we want to maximize area! This means that the x-value that we want should generate the larger of the areas when substituted into the area expression we defined earlier:

Area = -x^3 + 12x
*if x = -2*
Area = -(-2)^3 + 12(-2) = -(-8) - 24 = 8 - 24 = -16 <-- well, we know that isn't possible to have a negative area!! x = 2 must be our solution, let's plug it in:

Area = -x^3 + 12x
*if x = 2*
Area = -(2)^3 + 12(2) = -(8) + 24 = 16

So our solution is that the maximum area of the rectangle under this curve is 16 sq. units.

That corresponds to answer C.

Answer 2

oh okay thank you so much!