A pyramid has a height h and a square base with side x. What rate is the volume decreasing when the height is 120 meters and the base length is 150 meters?

Question

The question states that a pyramid has height h and a square base with side x. If the height decreases at a rate of .0005 meters per year and the side of the base is decreasing by .002 meters per year, what rate is the volume decreasing when the height is 120 meters and the absence length is 150 meters. 

We are learning related rate in class. This question I tried to complete the steps so I tried to draw a picture first. but after the picture I have no clue what to do. please help me. 

Answer 1

Volume of square pyramidis  V = pi*(x^2)*h/3

 

x is the side of base. h is the height.Here the simple logic behind the probelm is V,x,h are functions of t the time.

 

dx/dt=-0.002 mts/yr    dh/dt=0.0005 mts/yr   V=V = (pi/3)*(x^2)*h/3

/3

use product rule to determine the derivative of Volume V w.r.t to t.Remember x, h are function of t

 

dV/dt=(pi/3)i*[2*x*h*(dx/dt)+(x^2(*(dh/dt)].The subtitute the values dx/dt and dh/dt.These values are constant for this problem.If they also can function of t or x or h. 

 

      =-(pi/3)*[0.004xh+0.001x^2]

 

Just substitute the values x, h to get the dV/dt

Thank you