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One diagonal of a rhombus is decreasing at a rate of 7 centimeters per minute and the other diagonal of the rhombus is increasing at a rate of 10 centimeters per minute.
At a certain instant, the decreasing diagonal is 4 centimeters and the increasing diagonal is 6 centimeters.
What is the rate of change of the area of the rhombus at that instant (in square centimeters per minute)?
Choose 1 answer:
(A) -1
(B) 
16
(C) -16
(D) 1
The area of a rhombus with diagonals 
d_(1) and 
d_(2) is 
(d_(1)d_(2))/(2).

One diagonal of a rhombus is decreasing at a rate of 77 centimeters per minute and the other diagonal of the rhombus is increasing at a rate of 1010 centimeters per minute.\newlineAt a certain instant, the decreasing diagonal is 44 centimeters and the increasing diagonal is 66 centimeters.\newlineWhat is the rate of change of the area of the rhombus at that instant (in square centimeters per minute)?\newlineChoose 11 answer:\newline(A) 1-1\newline(B) 16 \mathbf{1 6} \newline(C) 16-16\newline(D) 11\newlineThe area of a rhombus with diagonals d1 d_{1} and d2 d_{2} is d1d22 \frac{d_{1} d_{2}}{2} .

Full solution

Q. One diagonal of a rhombus is decreasing at a rate of 77 centimeters per minute and the other diagonal of the rhombus is increasing at a rate of 1010 centimeters per minute.\newlineAt a certain instant, the decreasing diagonal is 44 centimeters and the increasing diagonal is 66 centimeters.\newlineWhat is the rate of change of the area of the rhombus at that instant (in square centimeters per minute)?\newlineChoose 11 answer:\newline(A) 1-1\newline(B) 16 \mathbf{1 6} \newline(C) 16-16\newline(D) 11\newlineThe area of a rhombus with diagonals d1 d_{1} and d2 d_{2} is d1d22 \frac{d_{1} d_{2}}{2} .
  1. Rhombus Area Formula: The formula for the area of a rhombus is (d1×d2)/2(d_1 \times d_2) / 2, where d1d_1 and d2d_2 are the lengths of the diagonals.
  2. Diagonal Lengths Given: Let's denote the decreasing diagonal as d1d_1 and the increasing diagonal as d2d_2. At the instant we are considering, d1=4cmd_1 = 4\,\text{cm} and d2=6cmd_2 = 6\,\text{cm}.
  3. Rates of Change: The rate of change of d1d_1 is 7cm/min-7\,\text{cm/min} (since it's decreasing) and the rate of change of d2d_2 is 10cm/min10\,\text{cm/min} (since it's increasing).
  4. Rate of Change Formula: To find the rate of change of the area, we'll use the product rule for differentiation, which in this context is d(Area)dt=(d1d(d2)dt+d2d(d1)dt)2\frac{d(\text{Area})}{dt} = \frac{(d_1 \cdot \frac{d(d_2)}{dt} + d_2 \cdot \frac{d(d_1)}{dt})}{2}.
  5. Calculation: Plugging in the values, we get d(Area)dt=(4cm×10cm/min+6cm×7cm/min)2\frac{d(\text{Area})}{dt} = \frac{(4 \, \text{cm} \times 10 \, \text{cm/min} + 6 \, \text{cm} \times -7 \, \text{cm/min})}{2}.
  6. Final Result: Now, let's do the calculation: d(Area)dt=(4042)2=22=1 cm2/min.\frac{d(\text{Area})}{dt} = \frac{(40 - 42)}{2} = \frac{-2}{2} = -1 \text{ cm}^2/\text{min}.

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