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The radius of a cylindrical box is 8 inches, and its height is 3 inches. Find the number of inches that you can add to the radius that would create the same increase in the volume of the box that would result if the same number of inches was instead added to the height.
something like this: >
v = pi *r^2 *h
let the number of inches added = x. then we want
pi* (r+x)^2 * h = pi *r^2*(h+x) ...so
-> hr^2 + 2hxr +hx^2= r^2*h + r^2*x
-> x*r^2- 2hxr-hx^2 =0= x(r^2-2hr-h^2) = x(r-h)^2
which implies that x =0
is this correct?
Using your dimensions (8, 3), I got expressions for two volumes, each involving a variable x.
Equating the volumes and simplifying resulted in the following.
x*(3*x - 16) = 0
Hence: x = 0 and x = 16/3 solve the problem with the dimensions you originally posted.