The length of a rectangle is increasing at a rate of 3 cm/s and its width is increasing at a rate of 8 cm/s. When the length is 15 cm and the width is 6 cm, how fast is the area of the rectangle increasing?

the area is increasing at 138 cm²/s

Step-by-step explanation:

Let L represent the length of the rectangle.

Let W represent the width of the rectangle

The formula for the area of the rectangle is expressed as

Area = LW

Since the length and width is increasing, we would apply the product rule to determine the rate at which the area is also increasing.

dy/dx = udv/dx + vdu/dx

L = u, W = v and x = t because it is changing with respect to time. Therefore,

dA/dt = Ldw/dt + Wdl/dt

From the information given,

L = 15cm

W = 6 cm

dw/dt = 8cm/s

dl/dt = 3cm/s

Therefore,

dA/dt = (15 × 8) + (6 × 3)

dA/dt = 120 + 18

dA/dt = 138 cm²/s

Given :

The length of a rectangle is increasing at a rate of 3cm/s and its width is increasing at a rate of 8 cm/s.

When the length is 15 cm and the width is 6 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 = 15*8 + 6*3

dA/dt = 120 + 18 cm²/s

dA/dt = 138 cm²/s

2 5/8 is the final answer! : )

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

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