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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) -8.5
(B) 8.5
(C) -21.5
(D) 21.5

The base of a triangle is decreasing at a rate of 1313 millimeters per minute and the height of the triangle is increasing at a rate of 66 millimeters per minute.\newlineAt a certain instant, the base is 55 millimeters and the height is 11 millimeter.\newlineWhat is the rate of change of the area of the triangle at that instant (in square millimeters per minute)?\newlineChoose 11 answer:\newline(A) 8-8.55\newline(B) 88.55\newline(C) 21-21.55\newline(D) 2121.55

Full solution

Q. The base of a triangle is decreasing at a rate of 1313 millimeters per minute and the height of the triangle is increasing at a rate of 66 millimeters per minute.\newlineAt a certain instant, the base is 55 millimeters and the height is 11 millimeter.\newlineWhat is the rate of change of the area of the triangle at that instant (in square millimeters per minute)?\newlineChoose 11 answer:\newline(A) 8-8.55\newline(B) 88.55\newline(C) 21-21.55\newline(D) 2121.55
  1. Area Formula: The area of a triangle is given by the formula A=12×base×heightA = \frac{1}{2} \times \text{base} \times \text{height}.
  2. Derivative Calculation: Let's denote the base by bb and the height by hh. The rate of change of the area with respect to time tt can be found using the derivative dAdt=12(dbdth+bdhdt)\frac{dA}{dt} = \frac{1}{2} \cdot (\frac{db}{dt} \cdot h + b \cdot \frac{dh}{dt}).
  3. Given Rates: Given dbdt=13\frac{db}{dt} = -13 mm/min (since the base is decreasing) and dhdt=6\frac{dh}{dt} = 6 mm/min (since the height is increasing).
  4. Instant Values: At the instant when the base bb is 5mm5\,\text{mm} and the height hh is 1mm1\,\text{mm}, we plug these values into the derivative formula to find dAdt\frac{dA}{dt}.
  5. Calculate dAdt\frac{dA}{dt}: So, dAdt=12×(13×1+5×6)\frac{dA}{dt} = \frac{1}{2} \times (-13 \times 1 + 5 \times 6).
  6. Final Result: Calculating the above expression gives dAdt=12×(13+30)\frac{dA}{dt} = \frac{1}{2} \times (-13 + 30).
  7. Final Result: Calculating the above expression gives dAdt=12×(13+30)\frac{dA}{dt} = \frac{1}{2} \times (-13 + 30). dAdt=12×17=8.5\frac{dA}{dt} = \frac{1}{2} \times 17 = 8.5 square millimeters per minute.

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