Consider the region R=(x,y):x y 9-(11)/(3)x^2,x 0.

Question

Consider the region $R={(x,y):x\le y\le 9-\frac{11}{3}{x}^{2},x\ge 0}$.
The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R , is:

Accepted Answer

$\mathrm{t} .\left(9-\frac{11 \mathrm{t}^2}{3}-\mathrm{t}\right)$

$$\begin{aligned}
& A=9 t-t^2-\frac{11}{3} t^3 \
& \frac{d A}{d t}=9-2 t-11 t^2 \
& \Rightarrow 11 t^2+2 t-9=0 \
& 11 t^2+11 t-9 t-9=0 \
& t=-1 & t=\frac{9}{11}
\end{aligned}$$

$\therefore \frac{\mathrm{dA}}{\mathrm{dt}}=$

$$\begin{aligned}
& \therefore \text { largest area }=\frac{9}{11}\left(9-\frac{11}{3}, \frac{81}{121}-\frac{9}{11}\right) \
& =\frac{9}{11} \cdot \frac{63}{11}=\frac{567}{121}
\end{aligned}$$

Solutions · 1