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The length of a rectangle is increasing at a rate of 8 cm/s and its width is increasing at a rate of 6 cm/s. when the length is 13 cm and the width is 5 cm, how fast is the area of the rectangle increasing?

Respuesta :

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L(t) = 13 + 8t ............(1)
W(t) = 5 + 6t .............(2)
Let the area be A.
A(t) = (13 + 8t)(5 + 6t) ...(3)
Expanding the brackets in (3), we get:
[tex]A(t)=48t^{2}+118t+65\ ..........(4)[/tex]
Differentiating A with respect to time gives:
 [tex]\frac{dA}{dt}=96t+118[/tex]
When t = 0, the rate of change of the area of the rectangle is: [tex]\frac{dA}{dt}=118[/tex]
Therefore the answer is 118 square cm per second.

VirαtKσhli

Given :

The length of a rectangle is increasing at a rate of 8 cm/s and its width is increasing at a rate of 6 cm/s.

When the length is 13 cm and the width is 5 cm .

To find :-

how fast is the area of the rectangle increasing?

Solution :-

As we know that :-

A = lb

To find the rate :-

d(A)/dt = d(lb)/dt .

Differenciate :-

dA/dt = l (db/dt ) + b (dl/dt )

Substitute :-

dA/dt = 13*7 + 5* 8

dA/dt = 91 + 40 cm²/s

dA/dt = 131 cm²/s

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