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) $148$ $\newline$ (B) $-204$ $\newline$ (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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Grade 8

Interpreting Linear Expressions

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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)

148

\newline

(B)

-204

\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

Formula Explanation: 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.
Chain Rule Application: We need to find the rate of change of the surface area with respect to time, which is $\frac{d(SA)}{dt}$ .
Derivative Calculations: To find $\frac{d(SA)}{dt}$ , we'll use the chain rule from calculus: $\frac{d(SA)}{dt} = \frac{d(SA)}{ds} \cdot \frac{ds}{dt} + \frac{d(SA)}{dh} \cdot \frac{dh}{dt}$ .
Rate of Change Calculation: First, we calculate $\frac{d(SA)}{ds} = 4s + 4h$ , since the derivative of $s^2$ with respect to $s$ is $2s$ and the derivative of $sh$ with respect to $s$ is $h$ .
Substitution of Values: Next, we calculate $\frac{d(SA)}{dh} = 4s$ , since the derivative of $sh$ with respect to $h$ is $s$ and the derivative of $s^2$ with respect to $h$ is $0$ .
Final Calculation: Now we plug in the values for $\frac{ds}{dt}$ and $\frac{dh}{dt}$ , which are $-7$ km/min and $10$ km/min, respectively.
Final Calculation: Now we plug in the values for $\frac{ds}{dt}$ and $\frac{dh}{dt}$ , which are $-7\,\text{km/min}$ and $10\,\text{km/min}$ , respectively.We also plug in the values for $s$ and $h$ at the given instant, which are $4\,\text{km}$ and $9\,\text{km}$ , respectively.
Final Calculation: Now we plug in the values for $\frac{ds}{dt}$ and $\frac{dh}{dt}$ , which are $-7 \, \text{km/min}$ and $10 \, \text{km/min}$ , respectively. We also plug in the values for $s$ and $h$ at the given instant, which are $4 \, \text{km}$ and $9 \, \text{km}$ , respectively. So, $\frac{d(SA)}{dt} = (4s + 4h)(-7) + (4s)(10)$ .
Final Calculation: Now we plug in the values for $\frac{ds}{dt}$ and $\frac{dh}{dt}$ , which are $-7 \, \text{km/min}$ and $10 \, \text{km/min}$ , respectively. We also plug in the values for $s$ and $h$ at the given instant, which are $4 \, \text{km}$ and $9 \, \text{km}$ , respectively. So, $\frac{d(SA)}{dt} = (4s + 4h)(-7) + (4s)(10)$ . Substitute $s = 4$ and $\frac{dh}{dt}$ $0$ into the equation: $\frac{dh}{dt}$ $1$ .
Final Calculation: Now we plug in the values for $\frac{ds}{dt}$ and $\frac{dh}{dt}$ , which are $-7 \, \text{km/min}$ and $10 \, \text{km/min}$ , respectively. We also plug in the values for $s$ and $h$ at the given instant, which are $4 \, \text{km}$ and $9 \, \text{km}$ , respectively. So, $\frac{d(SA)}{dt} = (4s + 4h)(-7) + (4s)(10)$ . Substitute $s = 4$ and $\frac{dh}{dt}$ $0$ into the equation: $\frac{dh}{dt}$ $1$ . Simplify the equation: $\frac{dh}{dt}$ $2$ .
Final Calculation: Now we plug in the values for $\frac{ds}{dt}$ and $\frac{dh}{dt}$ , which are $-7 \, \text{km/min}$ and $10 \, \text{km/min}$ , respectively. We also plug in the values for $s$ and $h$ at the given instant, which are $4 \, \text{km}$ and $9 \, \text{km}$ , respectively. So, $\frac{d(SA)}{dt} = (4s + 4h)(-7) + (4s)(10)$ . Substitute $s = 4$ and $\frac{dh}{dt}$ $0$ into the equation: $\frac{dh}{dt}$ $1$ . Simplify the equation: $\frac{dh}{dt}$ $2$ . Calculate the values: $\frac{dh}{dt}$ $3$ .
Final Calculation: Now we plug in the values for $\frac{ds}{dt}$ and $\frac{dh}{dt}$ , which are $-7 \, \text{km/min}$ and $10 \, \text{km/min}$ , respectively. We also plug in the values for $s$ and $h$ at the given instant, which are $4 \, \text{km}$ and $9 \, \text{km}$ , respectively. So, $\frac{d(SA)}{dt} = (4s + 4h)(-7) + (4s)(10)$ . Substitute $s = 4$ and $\frac{dh}{dt}$ $0$ into the equation: $\frac{dh}{dt}$ $1$ . Simplify the equation: $\frac{dh}{dt}$ $2$ . Calculate the values: $\frac{dh}{dt}$ $3$ . $\frac{dh}{dt}$ $4$ .
Final Calculation: Now we plug in the values for $\frac{ds}{dt}$ and $\frac{dh}{dt}$ , which are $-7 \, \text{km/min}$ and $10 \, \text{km/min}$ , respectively. We also plug in the values for $s$ and $h$ at the given instant, which are $4 \, \text{km}$ and $9 \, \text{km}$ , respectively. So, $\frac{d(SA)}{dt} = (4s + 4h)(-7) + (4s)(10)$ . Substitute $s = 4$ and $\frac{dh}{dt}$ $0$ into the equation: $\frac{dh}{dt}$ $1$ . Simplify the equation: $\frac{dh}{dt}$ $2$ . Calculate the values: $\frac{dh}{dt}$ $3$ . $\frac{dh}{dt}$ $4$ . Finally, $\frac{dh}{dt}$ $5$ square kilometers per minute.