Solve the problem.
-A satellite camera takes a rectangular-shaped picture. The smallest region that can be photographed is a $4 - \mathrm { km }$ by 6-km rectangle. As the camera zooms out, the length 1 and width $w$ of the rectangle increase at a rate of 3 $\mathrm { km } / \mathrm { sec }$. How long does it take for the area A to be at least 4 times its original size?
A) $1.61 \mathrm { sec }$
B) $9.7 \mathrm { sec }$
C) $4.94 \mathrm { sec }$
D) $3.28 \mathrm { sec }$

Question

Quizplus · Accepted Answer

The area of the rectangle is $A = lw$. The rate of change of the area is $$\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt} = 6(3) + 4(3) = 30 \mathrm { km }^2 / \mathrm { sec }$$The original area is $A_0 = 4 \cdot 6 = 24 \mathrm { km }^2$. We want to find the time $t$ when $A = 4A_0 = 96 \mathrm { km }^2$.$$\begin{split} \frac{dA}{dt} &= 30 \\ A - A_0 &= 30t \\ 96 - 24 &= 30t \\ t &= \frac{72}{30} = 2.4 \mathrm { sec } \end{split}$$Therefore, the answer is (A).

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