Posted by **<3** on Thursday, June 4, 2009 at 10:07pm.

The altitude of a triangle is increasing at a rate of 2.500 centimeters/minute while the area of the triangle is increasing at a rate of 1.500 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 10.500 centimeters and the area is 93.000 square centimeters?

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**MathMate**, Thursday, June 4, 2009 at 10:35pm
Given

A=bh/2, b=2*93/10.5=17.714

differentiate with respect to t (by the product rule on the right hand side).

The only unknown left is the rate of change of the base.

Hint: it is negative.

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**Reiny**, Thursday, June 4, 2009 at 10:38pm
let the area be A

let the base be x

let the height be y

given: at a time of t minutes,

dA/dt = 1.5 cm^2/min

dy/dt = 2.5 cm^2/min

find:

dx/dt when A = 93 and y = 10.5

A = xy/2 or

2A = xy (equ#1)

differentiate implicitly with respect to t, using the product rule

2dA/dt =x(dy/dt) + y(dx/dt) (equ#2)

we know A=93 when y = 10.5, so x = 17.714

sub into equ#2

2(1.5) = 17.714(2.5) + 10.5dx/dt

solve for dx/dt

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**<3**, Thursday, June 4, 2009 at 10:54pm
My answer is -3.9319, but it's wrong.

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**MathMate**, Thursday, June 4, 2009 at 10:58pm
I have -3.932 too.

Can you check the numbers in the question?

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**<3**, Thursday, June 4, 2009 at 11:01pm
I copied & pasted this from my online homework. It said the answer is wrong.

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**Reiny**, Thursday, June 4, 2009 at 11:02pm
I also had -3.932

What answer did the book have?

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**<3**, Thursday, June 4, 2009 at 11:07pm
I use an online homework program called WebWork. It just tells you if your answer is correct or not.

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**MathMate**, Thursday, June 4, 2009 at 11:10pm
Try -3.93, -3.932.

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**<3**, Thursday, June 4, 2009 at 11:16pm
The answer is -3.9300. Thanks!

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