The radius $R$ of a cylindrical box is $8$ cm, the height $H$ is $3$ cm. The volume $V = \pi R^{2}H$ is to be increased by the same fixed positive amount when $R$ is increased by $x$ inches as when $H$ is increased by $x$ cm. This condition is satisfied by

The initial volume of the cylinder is π(8^2)(3) = 192π. Increasing the height by x gives a volume increase of π(8^2)(x) = 64πx. Increasing the radius by x gives a new volume of π(8 + x)^2(3), so the volume increase is 3π(64 + 16x + x^2) - 192π = 48πx + 3πx^2. Setting the increases equal gives 64πx = 48πx + 3πx^2, which simplifies to 3x^2 = 16x. Since x must be positive, x = 16/3, which is one rational, but not integral, value of x.

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