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the length of a rectangle is increasing at a rate of 3 cm/s and its width is increasing at a rate of 7 cm/s. when the length is 15 cm and the width is 8 cm, how fast is the area of the rectangle increasing (in cm2/s)?

User Manki
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To find the rate at which the area of the rectangle is increasing, we can use the formula for the area of a rectangle, which is A = length × width.

Let's denote:

- the length of the rectangle as L(t)

- the width of the rectangle as W(t)

- the area of the rectangle as A(t)

Given that the length is increasing at a rate of 3 cm/s and the width is increasing at a rate of 7 cm/s, we can express these rates as the derivatives dL/dt and dW/dt, where t represents time.

We can also express the area A as a function of time: A(t) = L(t) × W(t).

To find the rate at which the area A is changing with respect to time t (dA/dt), we'll use the product rule of differentiation.

The product rule states:

If y = u × v, where u and v are functions of t, then

dy/dt = u × dv/dt + v × du/dt.

Applying the product rule to our scenario:

A'(t) = L(t) × dW/dt + W(t) × dL/dt

Given that the length is 15 cm and the width is 8 cm, we can find the values of dL/dt and dW/dt by substituting these values into the equation.

A'(t) = 15 × 7 + 8 × 3

A'(t) = 105 + 24

A'(t) = 129 cm^2/s

Therefore, when the length is 15 cm and the width is 8 cm, the area of the rectangle is increasing at a rate of 129 cm^2/s.

User Pemba Tamang
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