One base of a trapezoid is decreasing at a rate of 8 kilometers per second and the height of the trapezoid is increasing at a rate of 5 kilometers per second. The other base of the trapezoid is fixed at 4 kilometers. At a certain instant, the decreasing base is 12 kilometers and the height is 2 kilometers. What is the rate of change of the area of the trapezoid at that instant (in square kilometers per second)? Choose 1 answer: (A) -32 (B) -22 (C) 22 (D) 32

Write Formula: First, let's write down the formula for the area of a trapezoid, which is A = (1/2) * (b1 + b2) * h, where b1 and b2 are the lengths of the two bases and h is the height.Plug in Values: Now, we plug in the values we know. At the instant we're considering, b1 is decreasing and is 12 km, b2 is fixed at 4 km, and h is increasing and is 2 km.Differentiate Area Formula: To find the rate of change of the area, we need to differentiate the area formula with respect to time (t). So, dA/dt = (1/2) * (db1/dt + db2/dt) * h + (1/2) * (b1 + b2) * dh/dt.Substitute Rates: We know db1/dt = -8 km/s (since b1 is decreasing), db2/dt = 0 km/s (since b2 is fixed), and dh/dt = 5 km/s.Calculate Result: Substitute these rates into the differentiation formula: dA/dt = (1/2) * (-8 + 0) * 2 + (1/2) * (12 + 4) * 5.Simplify Equation: Now, let's do the math: dA/dt = (1/2) * (-8) * 2 + (1/2) * 16 * 5.Simplify the equation: dA/dt = -8 + 40.

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One base of a trapezoid is decreasing at a rate of $8$ kilometers per second and the height of the trapezoid is increasing at a rate of $5$ kilometers per second. $\newline$ The other base of the trapezoid is fixed at $4$ kilometers. $\newline$ At a certain instant, the decreasing base is $12$ kilometers and the height is $2$ kilometers. $\newline$ What is the rate of change of the area of the trapezoid at that instant (in square kilometers per second)? $\newline$ Choose $1$ answer: $\newline$ (A) $-32$ $\newline$ (B) $-22$ $\newline$ (C) $22$ $\newline$ (D) $32$

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Q. One base of a trapezoid is decreasing at a rate of

8

kilometers per second and the height of the trapezoid is increasing at a rate of

5

kilometers per second.

\newline

The other base of the trapezoid is fixed at

4

kilometers.

\newline

At a certain instant, the decreasing base is

12

kilometers and the height is

2

kilometers.

\newline

What is the rate of change of the area of the trapezoid at that instant (in square kilometers per second)?

\newline

Choose

1

answer:

\newline

(A)

-32

\newline

(B)

-22

\newline

(C)

22

\newline

(D)

32

Write Formula: First, let's write down the formula for the area of a trapezoid, which is $A = (\frac{1}{2}) \cdot (b_1 + b_2) \cdot h$ , where $b_1$ and $b_2$ are the lengths of the two bases and $h$ is the height.
Plug in Values: Now, we plug in the values we know. At the instant we're considering, $b_1$ is decreasing and is $12\,\text{km}$ , $b_2$ is fixed at $4\,\text{km}$ , and $h$ is increasing and is $2\,\text{km}$ .
Differentiate Area Formula: To find the rate of change of the area, we need to differentiate the area formula with respect to time $t$ . So, $\frac{dA}{dt} = \frac{1}{2} \cdot (\frac{db_1}{dt} + \frac{db_2}{dt}) \cdot h + \frac{1}{2} \cdot (b_1 + b_2) \cdot \frac{dh}{dt}$ .
Substitute Rates: We know $\frac{db_1}{dt} = -8 \, \text{km/s}$ (since $b_1$ is decreasing), $\frac{db_2}{dt} = 0 \, \text{km/s}$ (since $b_2$ is fixed), and $\frac{dh}{dt} = 5 \, \text{km/s}$ .
Calculate Result: Substitute these rates into the differentiation formula: $\frac{dA}{dt} = \left(\frac{1}{2}\right) \cdot (-8 + 0) \cdot 2 + \left(\frac{1}{2}\right) \cdot (12 + 4) \cdot 5$ .
Simplify Equation: Now, let's do the math: $\frac{dA}{dt} = \left(\frac{1}{2}\right) \cdot (-8) \cdot 2 + \left(\frac{1}{2}\right) \cdot 16 \cdot 5$ .
Simplify Equation: Now, let's do the math: $\frac{dA}{dt} = \left(\frac{1}{2}\right) \cdot (-8) \cdot 2 + \left(\frac{1}{2}\right) \cdot 16 \cdot 5$ . Simplify the equation: $\frac{dA}{dt} = -8 + 40$ .