Related rates problem: the radius of a spherical watermelon is growing at a constant rate of 2 centimeters per week. the thickness of the rind is always one tenth of the radius. the volume of the rind is growing at the rate cubic centimeters per week at the end of the fifth week. assume that the radius is initially zero.

 The volume of a sphere is V = 4/3 x pi x r^3 

The rind volume is the difference of two spheres. The radius of the inner sphere volume is, r - .1 r or .9 r. So the volume of the rind is 

V = 4/3 x pi x r^3 - 4/3 x pi (.9 r )^3 

V = 4/3 x pi x r^3 - 4/3 x pi x (.9^3) x r^3 

Implicitly differentiating both sides with respect to t yields: 

V' = ( 4 x pi x r^2 x r' ) - ( 4 x pi x (.9^3) x r^2 x r' ) 

This equation define the rate of change of the rind volume with respect to time. 

As defined in the problem, r' = 2 

As well, after 5 weeks the radius will be r = 2 * 5 = 10 

V' = (4 x pi x 10^2 x 2) - (4 x pi x (.9^3) x 10 x 2) 

V' = (800 x pi) - (800 x pi x (.729)) 

V' = (800 x pi) (1 - .729) 

V' = (800 x pi)(.271) 

V' = 681.09 cubic centimeters

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D. i think it is the savings

Answer: c. x y2 04 66 16