The side of the base of a square prism is decreasing at a rate of $7$ kilometers per minute and the height of the prism is increasing at a rate of $10$ kilometers per minute. $\newline$ At a certain instant, the base's side is $4$ kilometers and the height is $9$ kilometers. $\newline$ What is the rate of change of the surface area of the prism at that instant (in square kilometers per minute)? $\newline$ Choose $1$ answer: $\newline$ (A) $-204$ $\newline$ (B) $-148$ $\newline$ (C) $148$ $\newline$ (D) $204$ $\newline$ The surface area of a square prism with base side $s$ and height $h$ is $2 s^{2}+4 s h$ .

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Q. The side of the base of a square prism is decreasing at a rate of

7

kilometers per minute and the height of the prism is increasing at a rate of

10

kilometers per minute.

\newline

At a certain instant, the base's side is

4

kilometers and the height is

9

kilometers.

\newline

What is the rate of change of the surface area of the prism at that instant (in square kilometers per minute)?

\newline

Choose

1

answer:

\newline

(A)

-204

\newline

(B)

-148

\newline

(C)

148

\newline

(D)

204

\newline

The surface area of a square prism with base side

s

and height

h

2 s^{2}+4 s h

Surface Area Formula: The surface area (SA) of a square prism is given by the formula $SA = 2s^2 + 4sh$ , where $s$ is the side of the base and $h$ is the height.
Differentiation with Respect to Time: To find the rate of change of the surface area, we need to differentiate the surface area formula with respect to time $t$ . So we get $\frac{d(SA)}{dt} = \frac{d(2s^2)}{dt} + \frac{d(4sh)}{dt}$ .
Product Rule Application: Using the product rule for differentiation, the rate of change of the first term $2s^2$ with respect to time is $\frac{d(2s^2)}{dt} = 4s \cdot \frac{ds}{dt}$ , since the derivative of $s^2$ with respect to $s$ is $2s$ and then we multiply by $\frac{ds}{dt}$ (the rate of change of $s$ ).
Rate of Change Calculation: Similarly, the rate of change of the second term $4sh$ with respect to time is $\frac{d(4sh)}{dt} = 4s \frac{dh}{dt} + 4h \frac{ds}{dt}$ , using the product rule (derivative of the first times the second plus the first times the derivative of the second).
Given Rates and Values: Now we plug in the rates given in the problem: $\frac{ds}{dt} = -7 \, \text{km/min}$ (since the side is decreasing) and $\frac{dh}{dt} = 10 \, \text{km/min}$ (since the height is increasing).
Calculation of First Term Rate: We also plug in the values for $s$ and $h$ at the instant given: $s = 4$ km and $h = 9$ km.
Calculation of Second Term Rate: So the rate of change of the first term $2s^2$ is $4s \cdot \frac{ds}{dt} = 4 \cdot 4 \, \text{km} \cdot (-7 \, \text{km/min}) = -112 \, \text{km}^2/\text{min}.$
Total Rate of Change Calculation: And the rate of change of the second term $4sh$ is $4s \cdot \frac{dh}{dt} + 4h \cdot \frac{ds}{dt} = 4 \cdot 4 \, \text{km} \cdot 10 \, \text{km/min} + 4 \cdot 9 \, \text{km} \cdot (-7 \, \text{km/min}) = 160 \, \text{km}^2/\text{min} - 252 \, \text{km}^2/\text{min}$ .
Total Rate of Change Calculation: And the rate of change of the second term $4sh$ is $4s \cdot \frac{dh}{dt} + 4h \cdot \frac{ds}{dt} = 4 \cdot 4 \, \text{km} \cdot 10 \, \text{km/min} + 4 \cdot 9 \, \text{km} \cdot (-7 \, \text{km/min}) = 160 \, \text{km}^2/\text{min} - 252 \, \text{km}^2/\text{min}$ . Adding these two rates of change together gives us the total rate of change of the surface area: $\frac{d(SA)}{dt} = -112 \, \text{km}^2/\text{min} + (160 \, \text{km}^2/\text{min} - 252 \, \text{km}^2/\text{min})$ .
Total Rate of Change Calculation: And the rate of change of the second term $4sh$ is $4s \cdot \frac{dh}{dt} + 4h \cdot \frac{ds}{dt} = 4 \cdot 4 \, \text{km} \cdot 10 \, \text{km/min} + 4 \cdot 9 \, \text{km} \cdot (-7 \, \text{km/min}) = 160 \, \text{km}^2/\text{min} - 252 \, \text{km}^2/\text{min}$ . Adding these two rates of change together gives us the total rate of change of the surface area: $\frac{d(SA)}{dt} = -112 \, \text{km}^2/\text{min} + (160 \, \text{km}^2/\text{min} - 252 \, \text{km}^2/\text{min})$ . Calculating the sum gives us $\frac{d(SA)}{dt} = -112 \, \text{km}^2/\text{min} - 92 \, \text{km}^2/\text{min} = -204 \, \text{km}^2/\text{min}$ .