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Class 12
Mathematics
Application of Derivatives
the length x of a rectangle is decreasing at the r...

Question:
The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

Answer:

Since the length (x) is decreasing at the rate of 5 cm/minute and the width (y) is increasing at the rate of 4 cm/minute, we have:

\begin{align} \frac{dx}{dt} = -5 \;cm/min\; and \; \frac{dy}{dt}= 4 \;cm/min\end{align}

(a) The perimeter (P) of a rectangle is given by,

P = 2(x + y)

\begin{align} \therefore\frac{dp}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right)= 2(-5+4)=-2\;cm/min\end{align}

Hence, the perimeter is decreasing at the rate of 2 cm/min.

(b) The area (A) of a rectangle is given by,

A = x⋅ y

\begin{align} \therefore\frac{dA}{dt} = \frac{dx}{dt}.y + x.\frac{dy}{dt}=-5y + 4x \end{align}

When x = 8 cm and y = 6 cm,

\begin{align} \frac{dA}{dt} = (-5 \times 6 + 4 \times 8)\; cm^2/min = 2\; cm^2/min\end{align}

Hence, the area of the rectangle is increasing at the rate of 2 cm²/min.

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Question:
The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

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Question: The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

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Question:
The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.