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What is the maximum area of a rectangle enclosed between the parabola  
\[y=16-x^2\]
  and the  
\[x\]
  axis?
From the diagram, the base of the rectangle is  
\[3x\]
  and the height is  
\[16-x^2\]
  os the area is  
\[A=x(16-x^2)=16x-x^3\]
.
\[\frac{dA}{dx}=16-3x^2\]
.
Set  
\[\frac{dA}{dx}=0\]
  and solve to give  
\[x= \frac{4 \sqrt{3}}{3}\]
.
The area is then  
\[A=16 \frac{4 \sqrt{3}}{3} - (\frac{4 \sqrt{3}}{3})^3= \frac{128 \sqrt{3}}{9}\]
.