Solve the problem. Round your answer, if appropriate.
-The volume of a rectangular box with a square base remains constant at $500 \mathrm {~cm} ^ { 3 }$ as the area of the base increases at a rate of $6 \mathrm {~cm} ^ { 2 } / \mathrm { sec }$. Find the rate at which the height of the box is decreasing when each side of the base is $15 \mathrm {~cm}$ long. (Do not round your answer.)
A) $\frac { 8 } { 9 } \mathrm {~cm} / \mathrm { sec }$
B) $\frac { 2 } { 75 } \mathrm {~cm} / \mathrm { sec }$
C) $\frac { 20 } { 9 } \mathrm {~cm} / \mathrm { sec }$
D) $\frac { 8 } { 135 } \mathrm {~cm} / \mathrm { sec }$

Question

Quizplus · Accepted Answer

Let the length, width, and height of the box be $l$, $w$, and $h$, respectively. We are given that $lw\cdot h=500$ and $\frac{d}{dt}(lw)=6$, when $l=w=15$. Taking the derivative with respect to $t$ of both sides of $lw\cdot h=500$, we have:
\begin{align*}
\frac{d}{dt}(lw\cdot h) &= \frac{d}{dt}(500) \
\frac{d}{dt}(lw)\cdot h + lw\cdot \frac{d}{dt}(h) &= 0 \
\end{align*}
Since $lw\cdot h=500$, we know that $h=\frac{500}{lw}$. Plugging this in, we have:
\begin{align*}
\frac{d}{dt}(lw)\cdot \frac{500}{lw} + lw\cdot \frac{d}{dt}\left(\frac{500}{lw}ight) &= 0 \
\frac{d}{dt}(lw)\cdot \frac{500}{lw} - lw\cdot \frac{500}{l^2}\cdot \frac{dl}{dt} &= 0 \
6\cdot \frac{500}{225} - 225\cdot \frac{500}{225^2}\cdot \frac{dh}{dt} &= 0 \
\frac{dh}{dt} &= \boxed{\frac{8}{135}}\mathrm{~cm/s}
\end{align*}

Solve the Problem 500 cm3500 \mathrm {~cm} ^ { 3 }500 cm3 As the Area of the Base Increases at a Rate

Solve the Problem $500 \mathrm {~cm} ^ { 3 }$ As the Area of the Base Increases at a Rate