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The altitude (i.e., height) of a triangle is increasing at a rate of 1.5 cm/minute while the area of the triangle is increasing at a rate of 1.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 12 centimeters and the area is 88 square centimeters? The base is changing at cm/min.

Respuesta :

Answer:

-1.583 cm/min

Step-by-step explanation:

The area of a triangle is given by

A=BH/2

A is area, B is base, H is height

Differentiating both over time

[tex]\frac{dA}{dt} =\frac{dB}{dt}\frac{H}{2}  + \frac{dH}{dt}\frac{B}{2}[/tex]

we have that

[tex]\frac{dH}{dt}=1.5 cm/min[/tex]

[tex]\frac{dA}{dt}=1.5 sqcm/min[/tex]

When H=12 and A=88 then B is

[tex]88=\frac{12B}{2}[/tex]

B=14.66 cm

Therefore

[tex]1.5=\frac{dB}{dt}\frac{12}{2}  + 1.5*\frac{14.66}{2}[/tex]

[tex]\frac{dB}{dt}=-1.583 cm/min[/tex]

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