What is the area of the largest rectangle that could under the curve y = 12-x2and above the x-axis

What is the area of the largest rectangle that could under the curve y = 12-x2 and above the x-axis?

y = 12 - x²

Area, A = xy

A(x) = x(12 - x²)
A(x) = 12x - x³

Max. Area: A'(x) = 0:

A'(x) = 12 - 3x²

12 - 3x² = 0
- 3x² = - 12
x² = - 12 / - 3
x² = 4
x = √4
x = ± 2

y = 12 - (± 2)²
y = 12 - 4
y = 8

Domain: - 2 < x < 2, so

x-Dimension = 4

Largest Possible Area = 4(8) = 32 sq. units
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The largest rectangle under the curve would have vertices (x0,y0), (-x0,y0), (x0,0), and (-x0,0) with (x0,y0) some point on the curve. The area of this rectangle is 2*x0*y0 and we want to maximize this to maximize the rectangle area. Since y=12-x^2, the area of the rectangle is 2*x0*(12-x0^2)=24*x0-2*x0^3. This is of maximum value when its derivative with respect to x0 is 0, so d/dx0 of this is 24-6*x0^2=0, so x0=2, and thus y0=8, so yes, your answer x=2 and y=8 is correct.

x1= -2 to x2= 2 length = x2-x1=4
y1=0 y2=8 so breadth=y2-y1=8
area=8*4=32. Max area