Forty square inches of material is available to make a cylindrical; can of tuna and water, we need to find the dimensions of the can that will give the most volume

∵ S.A = 40 inches²

∵ S.A = 2π r h + 2π r²

∴ 2π r h + 2π r² = 40

Let us use this rule to find h in terms of r

- Subtract 2π r² from both sides

∵ 2π r h = 40 - 2 π r²

- Divide both sides by 2π r

∴ $h=\frac{40-2\pi r^{2}}{2\pi r}$

∴ $h=\frac{40}{2\pi r}-\frac{2\pi r^{2}}{2\pi r}$

∴ $h=\frac{20}{\pi r}-r$

∵ V = π r² h

- Substitute h by its value above

∴ $V=\pi r^{2}(\frac{20}{\pi r}-r)$

∴ V = 20 r - π r³

To find the most volume differentiate it with respect to r and equate it by 0 to find the value of r

∵ $\frac{dV}{dr}$ = 20 - 3π r²

∵ $\frac{dV}{dr}$ = 0

∴ 20 - 3π r² = 0

- Add 3π r² to both sides

∴ 20 = 3π r²

- Divide both sides by 3π

∴ r² = 2.122

- Take √ for both sides

∴ r = 1.457 inches

To find h substitute the value of r in the expression of h

∵ $h=\frac{20}{\pi r}-r$

∴ $h=\frac{20}{\pi (1.457)}-(1.457)$

∴ h = 2.913 inches

The dimensions of the can are 1.457 inches and 2.913 inches that will give the most volume

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