Solve the problem.
-Determine the dimensions of the rectangle of largest area that can be inscribed in a semicircle of radius $3 .$
A) $h = \sqrt { 2 } , w = 3 \sqrt { 2 }$
B) $h = 3 \sqrt { 2 } , w = \sqrt { 2 }$
C) $h = 3 \sqrt { 2 } , w = \frac { 3 \sqrt { 2 } } { 2 }$,
D) $h = \frac { 3 \sqrt { 2 } } { 2 } , w = 3 \sqrt { 2 }$

Question

Quizplus · Accepted Answer

Let the rectangle have height $h$ and width $w$. Then the diagonals of the rectangle are both radii of the semicircle, so $w^2+h^2=3^2=9$. The area of the rectangle is $A=wh$. Solving for $h$ in terms of $w$ in the equation $w^2+h^2=9$, we get $h=\sqrt{9-w^2}$. We can now express the area in terms of just $w$: $$A=w\sqrt{9-w^2}.$$ We want to maximize this function for $w\in[0,3]$. Taking the derivative with respect to $w$, we have $$\frac{dA}{dw}=\sqrt{9-w^2}-\frac{w^2}{\sqrt{9-w^2}}.$$ Setting this equal to $0$, we can solve for $w$ to find the value that maximizes the area. We get $w=\frac{3}{\sqrt{2}}$, so $h=\frac{3}{\sqrt{2}}$ as well. Therefore, the dimensions of the rectangle of maximum area are $w=3/\sqrt{2}$ and $h=3/\sqrt{2}$. The answer is choice D.

Solve the Problem 3.3 .3. A) h=2,w=32h = \sqrt { 2 } , w = 3 \sqrt { 2 }h=2​,w=32​

Solve the Problem $3 .$ A) $h = \sqrt { 2 } , w = 3 \sqrt { 2 }$