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

The side of the base of a square prism is decreasing at a rate of 77 kilometers per minute and the height of the prism is increasing at a rate of 1010 kilometers per minute.\newlineAt a certain instant, the base's side is 44 kilometers and the height is 99 kilometers.\newlineWhat is the rate of change of the surface area of the prism at that instant (in square kilometers per minute)?\newlineChoose 11 answer:\newline(A) 148148\newline(B) −204-204\newline(C) −148-148\newline(D) 204204\newlineThe surface area of a square prism with base side s s and height h h is 2s2+4sh 2 s^{2}+4 s h .

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Full solution

Q. The side of the base of a square prism is decreasing at a rate of 77 kilometers per minute and the height of the prism is increasing at a rate of 1010 kilometers per minute.\newlineAt a certain instant, the base's side is 44 kilometers and the height is 99 kilometers.\newlineWhat is the rate of change of the surface area of the prism at that instant (in square kilometers per minute)?\newlineChoose 11 answer:\newline(A) 148148\newline(B) −204-204\newline(C) −148-148\newline(D) 204204\newlineThe surface area of a square prism with base side s s and height h h is 2s2+4sh 2 s^{2}+4 s h .
  1. Formula Explanation: The surface area (SA) of a square prism is given by the formula SA=2s2+4shSA = 2s^2 + 4sh, where ss is the side of the base and hh is the height.
  2. Chain Rule Application: We need to find the rate of change of the surface area with respect to time, which is d(SA)dt\frac{d(SA)}{dt}.
  3. Derivative Calculations: To find d(SA)dt\frac{d(SA)}{dt}, we'll use the chain rule from calculus: d(SA)dt=d(SA)ds⋅dsdt+d(SA)dh⋅dhdt\frac{d(SA)}{dt} = \frac{d(SA)}{ds} \cdot \frac{ds}{dt} + \frac{d(SA)}{dh} \cdot \frac{dh}{dt}.
  4. Rate of Change Calculation: First, we calculate d(SA)ds=4s+4h\frac{d(SA)}{ds} = 4s + 4h, since the derivative of s2s^2 with respect to ss is 2s2s and the derivative of shsh with respect to ss is hh.
  5. Substitution of Values: Next, we calculate d(SA)dh=4s\frac{d(SA)}{dh} = 4s, since the derivative of shsh with respect to hh is ss and the derivative of s2s^2 with respect to hh is 00.
  6. Final Calculation: Now we plug in the values for dsdt\frac{ds}{dt} and dhdt\frac{dh}{dt}, which are −7-7 km/min and 1010 km/min, respectively.
  7. Final Calculation: Now we plug in the values for dsdt\frac{ds}{dt} and dhdt\frac{dh}{dt}, which are −7 km/min-7\,\text{km/min} and 10 km/min10\,\text{km/min}, respectively.We also plug in the values for ss and hh at the given instant, which are 4 km4\,\text{km} and 9 km9\,\text{km}, respectively.
  8. Final Calculation: Now we plug in the values for dsdt\frac{ds}{dt} and dhdt\frac{dh}{dt}, which are −7 km/min-7 \, \text{km/min} and 10 km/min10 \, \text{km/min}, respectively. We also plug in the values for ss and hh at the given instant, which are 4 km4 \, \text{km} and 9 km9 \, \text{km}, respectively. So, d(SA)dt=(4s+4h)(−7)+(4s)(10)\frac{d(SA)}{dt} = (4s + 4h)(-7) + (4s)(10).
  9. Final Calculation: Now we plug in the values for dsdt\frac{ds}{dt} and dhdt\frac{dh}{dt}, which are −7 km/min-7 \, \text{km/min} and 10 km/min10 \, \text{km/min}, respectively. We also plug in the values for ss and hh at the given instant, which are 4 km4 \, \text{km} and 9 km9 \, \text{km}, respectively. So, d(SA)dt=(4s+4h)(−7)+(4s)(10)\frac{d(SA)}{dt} = (4s + 4h)(-7) + (4s)(10). Substitute s=4s = 4 and dhdt\frac{dh}{dt}00 into the equation: dhdt\frac{dh}{dt}11.
  10. Final Calculation: Now we plug in the values for dsdt\frac{ds}{dt} and dhdt\frac{dh}{dt}, which are −7 km/min-7 \, \text{km/min} and 10 km/min10 \, \text{km/min}, respectively. We also plug in the values for ss and hh at the given instant, which are 4 km4 \, \text{km} and 9 km9 \, \text{km}, respectively. So, d(SA)dt=(4s+4h)(−7)+(4s)(10)\frac{d(SA)}{dt} = (4s + 4h)(-7) + (4s)(10). Substitute s=4s = 4 and dhdt\frac{dh}{dt}00 into the equation: dhdt\frac{dh}{dt}11. Simplify the equation: dhdt\frac{dh}{dt}22.
  11. Final Calculation: Now we plug in the values for dsdt\frac{ds}{dt} and dhdt\frac{dh}{dt}, which are −7 km/min-7 \, \text{km/min} and 10 km/min10 \, \text{km/min}, respectively. We also plug in the values for ss and hh at the given instant, which are 4 km4 \, \text{km} and 9 km9 \, \text{km}, respectively. So, d(SA)dt=(4s+4h)(−7)+(4s)(10)\frac{d(SA)}{dt} = (4s + 4h)(-7) + (4s)(10). Substitute s=4s = 4 and dhdt\frac{dh}{dt}00 into the equation: dhdt\frac{dh}{dt}11. Simplify the equation: dhdt\frac{dh}{dt}22. Calculate the values: dhdt\frac{dh}{dt}33.
  12. Final Calculation: Now we plug in the values for dsdt\frac{ds}{dt} and dhdt\frac{dh}{dt}, which are −7 km/min-7 \, \text{km/min} and 10 km/min10 \, \text{km/min}, respectively. We also plug in the values for ss and hh at the given instant, which are 4 km4 \, \text{km} and 9 km9 \, \text{km}, respectively. So, d(SA)dt=(4s+4h)(−7)+(4s)(10)\frac{d(SA)}{dt} = (4s + 4h)(-7) + (4s)(10). Substitute s=4s = 4 and dhdt\frac{dh}{dt}00 into the equation: dhdt\frac{dh}{dt}11. Simplify the equation: dhdt\frac{dh}{dt}22. Calculate the values: dhdt\frac{dh}{dt}33. dhdt\frac{dh}{dt}44.
  13. Final Calculation: Now we plug in the values for dsdt\frac{ds}{dt} and dhdt\frac{dh}{dt}, which are −7 km/min-7 \, \text{km/min} and 10 km/min10 \, \text{km/min}, respectively. We also plug in the values for ss and hh at the given instant, which are 4 km4 \, \text{km} and 9 km9 \, \text{km}, respectively. So, d(SA)dt=(4s+4h)(−7)+(4s)(10)\frac{d(SA)}{dt} = (4s + 4h)(-7) + (4s)(10). Substitute s=4s = 4 and dhdt\frac{dh}{dt}00 into the equation: dhdt\frac{dh}{dt}11. Simplify the equation: dhdt\frac{dh}{dt}22. Calculate the values: dhdt\frac{dh}{dt}33. dhdt\frac{dh}{dt}44. Finally, dhdt\frac{dh}{dt}55 square kilometers per minute.

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