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September 30, 2014

September 30, 2014

Posted by **Kitor** on Wednesday, May 22, 2013 at 10:18pm.

I have tried to answer this in so many ways but couldn't get right answer ... Help please !

- Math calculus -
**Reiny**, Wednesday, May 22, 2013 at 10:53pmLet the base be x cm and the height be y cm

Area = (1/2)xy cm^2

dArea/dt = (1/2)(x dy/dt + y dx/dt) ------- #1

given: dy/dt = 1000 cm/min

dArea/dt = 1500 cm^2/min

find dx/dt when y = 7000 cm and Area = 89000 cm^2

...........

when y = 7000 and area = 89000

(1/2)(x)(7000) = 89000

x = 178/7

which leaves dx/dt as the only missing part of #1

1500 = (1/2)( (178/7)(1000) + 7000(dx/dt) )

3000 = 178000/7 + 7000 dx/dt

dx/dt = -157000/49000 cm/min

= - 157/49 cm/min

= appr -3.2 cm/min

At that instant, the base is decreasing at a rate of appr 3.2 cm/min

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