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)?

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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)?

Find Surface Area: First, let's find the expression for the surface area of a square prism. A square prism has $2$ square bases and $4$ rectangular sides. The surface area (SA) is given by: $\newline$ $SA = 2 \times (\text{side length})^2 + 4 \times (\text{side length}) \times \text{height}$
Differentiate Surface Area: Now, let's differentiate the surface area with respect to time to find the rate of change of the surface area. We'll use the product rule for differentiation where necessary. $\newline$ $\frac{d(SA)}{dt} = 2 \times 2 \times (\text{side length}) \times \frac{d(\text{side length})}{dt} + 4 \times [\frac{d(\text{side length})}{dt} \times \text{height} + (\text{side length}) \times \frac{d(\text{height})}{dt}]$
Given Rates: We are given the rates of change of the side length and the height: $\newline$ $\frac{d(\text{side length})}{dt} = -7 \, \text{km/min}$ (since the side is decreasing) $\newline$ $\frac{d(\text{height})}{dt} = 10 \, \text{km/min}$ (since the height is increasing)
Substitute Values: Now we substitute the given values and the rates into the differentiated surface area formula: $\newline$ $\frac{d(SA)}{dt} = 2 \times 2 \times 4 \, \text{km} \times (-7 \, \text{km/min}) + 4 \times [-7 \, \text{km/min} \times 9 \, \text{km} + 4 \, \text{km} \times 10 \, \text{km/min}]$
Perform Calculations: Let's perform the calculations: $\newline$ $\frac{d(SA)}{dt} = 2 \times 2 \times 4 \times (-7) + 4 \times [-7 \times 9 + 4 \times 10]$ $\newline$ $\frac{d(SA)}{dt} = 16 \times (-7) + 4 \times [-63 + 40]$ $\newline$ $\frac{d(SA)}{dt} = -112 + 4 \times (-23)$ $\newline$ $\frac{d(SA)}{dt} = -112 - 92$ $\newline$ $\frac{d(SA)}{dt} = -204$ square kilometers per minute