Finding the Rate of Change of Area with Decreasing Length and Increasing Width

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W0 + 2L0 - 9tA'(t) = -3(8) + 2(15) - 9tA'(t) = -24 + 30 - 9tA'(t) = 6 - 9tIn summary, the question asks for the rate of change of the area of a rectangle with a decreasing length and increasing width. The formula for the area is A = W * L and using differentiation, the rate of change is found to be A'(t) = 6 - 9t.
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kioria
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Hi,

I was practising differentiation and came across a some what rather easy question which I could not do. Can anyone direct me as usual. (this place is becoming my e-home). Mind you I am enjoying using LaTex converters haha.

Q. The length L of a rectangle is decreasing at the rate of 3 cm/sec, and the width W is increasing at the rate of 2 cm/sec. Find the rate of change of the area when L = 15 cm and W = 8 cm.

Thanks.
 
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  • #2
dL/dt = -3 and dW/dt = +2. The area A is equal to W * L. Differentiate wrt t and plug in the values you were given.
 
  • #3
W(t) = W0 + 2t
L(t) = L0 - 3t

A(t) = W(t)L(t)
A'(t) = W(t)L'(t) + W'(t)L(t)
A'(t) = -3(W0 + 2t) + 2(L0 - 3t)
 

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