At what rate is the base of the triangle changing when the altitude is 8 centimeters and the area is 100 square centimeters?

Question

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

Answer 1

Harmoni,

A(t)=((1/2)*b(t)*h(t)

dA/dt=(1/2)*(b*dh/dt+h*db/dt)

where A=area, h=altitude and b=base (base = 2*100/8=25)

plugging in, 2.5=(1/2)*(25*2+8*db/dt)

db/dt=-45/8 cm/min.

Answer 2

Area A of triangle A = 1/2bh; dA/dt = 1/2hdb/dt + 1/2bdh/dt;

Let substitude given data: 2.5 = 1/2·8·db/dt + 1/2b·2;

Now let find base: 100 = 1/2·8·b; b = 25 cm;

2.5 = 4db/dt + 25; db/dt = -5.625 cm/min decreasing