Asked by lana
A rectangle is expanding such that the ratio of its width to its length is always 2 to 5. Determine the rate of increase of the area of the rectangle when the perimeter is increasing at 10 cm/s and the perimeter is 35 cm.
Answers
Answered by
mathhelper
Let the width be 2x
let the length be 5x
Perimeter = 14x
dP/dt = 14 dx/dt
given: dP/dt = 10 cm/s, when P = 35
when P = 35
14x = 35
x = 35/14 = 5/2 cm
area = (2x)(5x) = 10x^2
d(area)/dt = 20x dx/dt
for our case:
dP/dt = 14 dx/dt = 10
dx/dt = 10/14 = 5/7
d(area)/dt = 20x dx/dt
= 20 (5/2)(5/7) cm^2/s
= 250/7 cm^2/s
let the length be 5x
Perimeter = 14x
dP/dt = 14 dx/dt
given: dP/dt = 10 cm/s, when P = 35
when P = 35
14x = 35
x = 35/14 = 5/2 cm
area = (2x)(5x) = 10x^2
d(area)/dt = 20x dx/dt
for our case:
dP/dt = 14 dx/dt = 10
dx/dt = 10/14 = 5/7
d(area)/dt = 20x dx/dt
= 20 (5/2)(5/7) cm^2/s
= 250/7 cm^2/s
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