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The perimeter of a square is increasing at a rate of 5 meters per hour.
At a certain instant, the perimeter is 30 meters.
What is the rate of change of the area of the square at that instant (in square meters per hour)?
Choose 1 answer:
(A)
(5)/(4)
(B) 25
(C)
(75)/(4)
(D)
(25)/(16)

The perimeter of a square is increasing at a rate of $5$ meters per hour. $\newline$ At a certain instant, the perimeter is $30$ meters. $\newline$ What is the rate of change of the area of the square at that instant (in square meters per hour)? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{5}{4}$ $\newline$ (B) $25$ $\newline$ (C) $\frac{75}{4}$ $\newline$ (D) $\frac{25}{16}$

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Roots of rational numbers

Full solution

Q. The perimeter of a square is increasing at a rate of

5

meters per hour.

\newline

At a certain instant, the perimeter is

30

meters.

\newline

What is the rate of change of the area of the square at that instant (in square meters per hour)?

\newline

Choose

1

answer:

\newline

(A)

\frac{5}{4}

\newline

(B)

25

\newline

(C)

\frac{75}{4}

\newline

(D)

\frac{25}{16}

Calculate Perimeter Length: First, let's find the length of one side of the square using the perimeter. The perimeter $P$ of a square is $4$ times the length of one side $s$ , so $P = 4s$ .
Solve for Side Length: Given $P = 30$ meters, we can solve for $s$ : $30 = 4s$ , so $s = \frac{30}{4} = 7.5$ meters.
Find Area: Now, we know the area $A$ of a square is $s^2$ . So, $A = (7.5)^2 = 56.25$ square meters.
Perimeter Rate of Change: The rate of change of the perimeter is $5$ meters per hour. Since the perimeter is $4$ times the length of one side, the rate of change of the length of one side is $\frac{5}{4}$ meters per hour.
Area Rate of Change: To find the rate of change of the area, we use the derivative of $A$ with respect to $s$ , which is $\frac{dA}{ds} = 2s$ . Then we multiply this by the rate of change of $s$ , which is $\frac{ds}{dt}$ .
Final Rate of Change: So, the rate of change of the area is $\frac{dA}{dt} = \frac{dA}{ds} \cdot \frac{ds}{dt} = 2s \cdot \left(\frac{5}{4}\right)$ .
Final Rate of Change: So, the rate of change of the area is $\frac{dA}{dt} = \frac{dA}{ds} \cdot \frac{ds}{dt} = 2s \cdot \left(\frac{5}{4}\right)$ . Plugging in the value of $s$ , we get $\frac{dA}{dt} = 2 \cdot 7.5 \cdot \left(\frac{5}{4}\right) = 15 \cdot \left(\frac{5}{4}\right) = 18.75$ square meters per hour.

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