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The base of a triangle is decreasing at a rate of 13 millimeters per minute and the height of the triangle is increasing at a rate of 6 millimeters per minute.
At a certain instant, the base is 5 millimeters and the height is 1 millimeter.
What is the rate of change of the area of the triangle at that instant (in square millimeters per minute)?
Choose 1 answer:
(A) 21.5
(B) -8.5
(C) -21.5
(D) 8.5

The base of a triangle is decreasing at a rate of $13$ millimeters per minute and the height of the triangle is increasing at a rate of $6$ millimeters per minute. $\newline$ At a certain instant, the base is $5$ millimeters and the height is $1$ millimeter. $\newline$ What is the rate of change of the area of the triangle at that instant (in square millimeters per minute)? $\newline$ Choose $1$ answer: $\newline$ (A) $21$ . $5$ $\newline$ (B) $-8$ . $5$ $\newline$ (C) $-21$ . $5$ $\newline$ (D) $8$ . $5$

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Calculate unit rates with fractions

Full solution

Q. The base of a triangle is decreasing at a rate of

13

millimeters per minute and the height of the triangle is increasing at a rate of

6

millimeters per minute.

\newline

At a certain instant, the base is

5

millimeters and the height is

1

millimeter.

\newline

What is the rate of change of the area of the triangle at that instant (in square millimeters per minute)?

\newline

Choose

1

answer:

\newline

(A)

21

.

5

\newline

(B)

-8

.

5

\newline

(C)

-21

.

5

\newline

(D)

8

.

5

Calculate Initial Area: Now, let's calculate the initial area of the triangle with the given base and height. $\newline$ Area = $(5 \, \text{mm} \times 1 \, \text{mm}) / 2 = \frac{5}{2} \, \text{mm}^2 = 2.5 \, \text{mm}^2$ .
Find Rate of Change: To find the rate of change of the area, we'll use the formula for the derivative of the area with respect to time, $\frac{dA}{dt} = \frac{1}{2} \times (\frac{db}{dt} \times h + b \times \frac{dh}{dt})$ , where $\frac{db}{dt}$ is the rate of change of the base and $\frac{dh}{dt}$ is the rate of change of the height.
Plug in Values: We know $\frac{db}{dt} = -13 \, \text{mm/min}$ (since the base is decreasing) and $\frac{dh}{dt} = 6 \, \text{mm/min}$ (since the height is increasing). Let's plug these values into the formula. $\frac{dA}{dt} = \frac{1}{2} \times (-13 \, \text{mm/min} \times 1 \, \text{mm} + 5 \, \text{mm} \times 6 \, \text{mm/min})$ .
Perform Calculation: Now, let's do the calculation. $\newline$ $\frac{dA}{dt} = \frac{1}{2} * (-13 \, \text{mm}^2/\text{min} + 30 \, \text{mm}^2/\text{min}) = \frac{1}{2} * 17 \, \text{mm}^2/\text{min} = 8.5 \, \text{mm}^2/\text{min}$ .

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