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The length x of a rectangle is decreasing at the rate of 3 cm/ mint and the width y is increasing at the rate of 2cm/min. when x = 10cm and y = 6cm, find the ratio of change of (a) the perimeter (b) the area of the rectangle

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Solution: Let length of rectangle =x cm and width of rectangle =y cm Given the length is decreasing at the rate of 3 cm/ minutes i.e. is decreasing w.r.t time dtdx=−3 cm/min.......(1) as x is decreasing and the width y is increasing at the rate of 2 cm/min i.e. y is increasing w.r.t...
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Solution:

Let length of rectangle =x cm
and width of rectangle =y cm
 
Given the length is decreasing at the rate of 3 cm/ minutes i.e. is decreasing w.r.t time 
 
dtdx=−3 cm/min.......(1) as x is decreasing and the width y is increasing at the rate of 2 cm/min
i.e. y is increasing w.r.t time
dy/dt=2 cm/min.........(2)
 
(1) Let P be the perimeter of rectangle
=2(l+width)
P=2(x+y)
dp/dt=2,(dx/dt+dy/dt)
Given : dx/dt=−3,dy/dt=2
∴dp/dt=2(−3+2)=2×−1=−2 cm/min
∴ perimeter is decreasing at the rate of 2 cm/min
 
(2) Let A be the area of the rectangle 
A=l×width=xy⇒dA/dt=xdy/dt+ydtdx
⇒dA/dt=−3y+2x from (1) & (2)
dA/dt∣∣∣∣x=10,y=6=−3×6+2×10=−18+20=2
 
Since are is m cm2dA/dt=2 cm2/min
 
Hence are is increasing at the rate of 2 cm2/min
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