Consider the region $R=\left\{(x,y):x\le y\le9-\frac{11}3x^2,\ x\ge0\right\}$. The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in $R$, is:

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Take the rectangle with a corner at $(0,p)$, the lower-right corner $(p,p)$ on $y=x$ and the upper-right corner $\left(p,9-\frac{11}3p^2\right)$ on the parabola. Area $A=p\left(9-\frac{11}3p^2-p\right)=9p-p^2-\frac{11}3p^3$. $A'=9-2p-11p^2=0\Rightarrow p=\frac9{11}$. $A=\frac{81}{11}-\frac{81}{121}-\frac{243}{121}=\frac{567}{121}$. Answer (4).

Q. Consider the region $R=\left\{(x,y):x\le y\le9-\frac{11}3x^2,\ x\ge0\right\}$. The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in $R$, is:

Explanation

Reviewed by Dr. Abubacker

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