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The altitude (i.e., height) of a triangle is increasing at a rate of 2 cm/minute while the area of the triangle is increasing at a rate of 4.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 9.5 centimeters and the area is 92 square centimeters?

Respuesta :

bartdrinksmalk
The formula for the area of a triangle is
[tex]A=\frac{1}{2}bh[/tex]
[tex]\rightarrow\frac{dA}{dt}=\frac{1}{2}(b\frac{dh}{dt}+h\frac{db}{dt})[/tex]

We are given:
[tex]\frac{dh}{dt}=2[/tex]
[tex]\frac{dA}{dt}=4.5[/tex]
[tex]h=9.5[/tex]
[tex]A=92[/tex]

Plug in:
[tex]4.5=\frac{1}{2}(b*2+9.5\frac{db}{dt})[/tex]
We can find [tex]b[/tex] with:
[tex]A=\frac{1}{2}bh\rightarrowb=\frac{2A}{h}=\frac{2*92}{9.5}=\frac{184}{9.5}[/tex]
So [tex]4.5=\frac{1}{2}(2*\frac{184}{9.5}+9.5\frac{db}{dt})[/tex]
[tex]\rightarrow9=\frac{368}{9.5}+9.5\frac{db}{dt}[/tex]
[tex]\rightarrow9-\frac{368}{9.5}=9.5\frac{db}{dt}[/tex]
[tex]\rightarrow\frac{db}{dt}=\frac{9-\frac{368}{9.5}}{9.5}\approx-3.130[/tex]

So the base is decreasing at a rate of 3.130 cm/minute.

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