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The side of the base of a square pyramid is increasing at a rate of 6 meters per minute and the height of the pyramid is decreasing at a rate of 1 meter per minute.
At a certain instant, the base's side is 3 meters and the height is 9 meters.
What is the rate of change of the volume of the pyramid at that instant (in cubic meters per minute)?
Choose 1 answer:
(A) -111
(B)
111
(C) -105
(D) 105
The volume of a square pyramid with base side
s and height
h is
(1)/(3)s^(2)h.

The side of the base of a square pyramid is increasing at a rate of $6$ meters per minute and the height of the pyramid is decreasing at a rate of $1$ meter per minute. $\newline$ At a certain instant, the base's side is $3$ meters and the height is $9$ meters. $\newline$ What is the rate of change of the volume of the pyramid at that instant (in cubic meters per minute)? $\newline$ Choose $1$ answer: $\newline$ (A) $-111$ $\newline$ (B) $\mathbf{1 1 1}$ $\newline$ (C) $-105$ $\newline$ (D) $105$ $\newline$ The volume of a square pyramid with base side $s$ and height $h$ is $\frac{1}{3} s^{2} h$ .

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Volume of cubes and rectangular prisms: word problems

Full solution

Q. The side of the base of a square pyramid is increasing at a rate of

6

meters per minute and the height of the pyramid is decreasing at a rate of

1

meter per minute.

\newline

At a certain instant, the base's side is

3

meters and the height is

9

meters.

\newline

What is the rate of change of the volume of the pyramid at that instant (in cubic meters per minute)?

\newline

Choose

1

answer:

\newline

(A)

-111

\newline

(B)

\mathbf{1 1 1}

\newline

(C)

-105

\newline

(D)

105

\newline

The volume of a square pyramid with base side

s

and height

h

is

\frac{1}{3} s^{2} h

.

Write Formula: First, let's write down the formula for the volume of a square pyramid, which is $V = \frac{1}{3}s^2h$ , where $s$ is the side length of the base and $h$ is the height.
Find Derivative: We need to find the rate of change of the volume, so we'll use calculus and take the derivative of the volume with respect to time, $\frac{dV}{dt}.$
Apply Values: The derivative of $V$ with respect to time $t$ is $\frac{dV}{dt} = \left(\frac{1}{3}\right)\left(2s\frac{ds}{dt}h + s^2\frac{dh}{dt}\right)$ , since we need to apply the product rule to $s^2h$ .
Calculate Result: We know $\frac{ds}{dt} = 6$ meters per minute (rate of change of the side) and $\frac{dh}{dt} = -1$ meter per minute (rate of change of the height, it's negative because the height is decreasing).
Calculate Result: We know $\frac{ds}{dt} = 6$ meters per minute (rate of change of the side) and $\frac{dh}{dt} = -1$ meter per minute (rate of change of the height, it's negative because the height is decreasing).Now we plug in the values of $s$ , $\frac{ds}{dt}$ , $h$ , and $\frac{dh}{dt}$ into the derivative. So, $\frac{dV}{dt} = (\frac{1}{3})(2\cdot3\cdot6\cdot9 + 3^2\cdot(-1))$ .
Calculate Result: We know $\frac{ds}{dt} = 6$ meters per minute (rate of change of the side) and $\frac{dh}{dt} = -1$ meter per minute (rate of change of the height, it's negative because the height is decreasing).Now we plug in the values of $s$ , $\frac{ds}{dt}$ , $h$ , and $\frac{dh}{dt}$ into the derivative. So, $\frac{dV}{dt} = (\frac{1}{3})(2\cdot3\cdot6\cdot9 + 3^2\cdot(-1))$ .Let's do the math: $\frac{dV}{dt} = (\frac{1}{3})(2\cdot3\cdot6\cdot9 + 9\cdot(-1)) = (\frac{1}{3})(324 - 9) = (\frac{1}{3})(315) = 105$ cubic meters per minute.

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