Find Maximum Area of Rectangle Inscribed Under Parabola $f(x)=-\frac{1}{3}x^2 +12$

Question

The parabola $f(x)=-\frac{1}{3}x^2 +12$ encloses an area with the x-axis. A rectangle is inscribed in this area. Which rectangle has the maximum area?

Accepted Answer

The rectangle has vertices at (6,0), (-6,0), (-6,6), (6,6); maximum area = 72

Answer

Step 1: Find x-intercepts of $f(x)$: set $-\frac{1}{3}x^2+12=0$ → $x^2=36$ → $x=6$ or $x=-6$. The rectangle has width $2x$ and height $f(x)$ Step 2: Define area function $A(x)=2x\cdot f(x)=2x(-\frac{1}{3}x^2+12)=-\frac{2}{3}x^3+24x$ Step 3: Find first derivative $A'(x)=-2x^2+24$, set to 0: $-2x^2+24=0$ → $x^2=12$ → $x=2\sqrt{3}$ (positive x only, since width is positive) Step 4: Calculate height $f(2\sqrt{3})=-\frac{1}{3}(12)+12=8$. Width = $4\sqrt{3}$. Area = $4\sqrt{3}\times8=32\sqrt{3}≈55.43$; or using symmetric vertex form, confirm maximum at $x=2\sqrt{3}$