Asked by Anonymous
if the length is decreasing at a rate of 2 in./min. while width is increasing at a rate of 2 in./min., what must be true about area of the rectangle
Answers
Answered by
Reiny
A = Lw
dA/dt = L dw/dt + w dL/dt
but dL/dt = -2
and dw/dt = 2
dA/dt = 2L - 2w
= 2(L-w)
So it depends on the size of the rectangle,
If the rectangle is a square (L=w), then area would remain constant,
if L > w, the area is increasing, and
if L < w ......
dA/dt = L dw/dt + w dL/dt
but dL/dt = -2
and dw/dt = 2
dA/dt = 2L - 2w
= 2(L-w)
So it depends on the size of the rectangle,
If the rectangle is a square (L=w), then area would remain constant,
if L > w, the area is increasing, and
if L < w ......
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