A cylindrical can without a top is to be made from 27cm^2
of sheet metal. Find the dimensions of the can with
the greatest volume. Be sure to formally introduce any new variables and their domains. Confirm you have a maximum
The area of the can will be "S =\\pi R^2 + 2\\pi R h," where R is radius of the bottom and h is the height.
The volume is "V = \\pi R^2 h" .
"S =\\pi R^2 + 2\\pi R h \\Rightarrow h = \\dfrac{S-\\pi R^2}{2\\pi R}, \\\\ V = \\pi R^2\\cdot \\dfrac{S-\\pi R^2}{2\\pi R} = 0.5(SR- \\pi R^3)."
Let us take the derivative with respect to R
"V' = 0.5S - 1.5\\pi R^2."
We should find a maximum, so we'll consider values for which the derivative is 0. Moreover, R should be positive.
"0.5S - 1.5\\pi R^2 = 0, \\;\\; R = \\sqrt{\\dfrac{0.5S}{1.5\\pi}} = \\sqrt{\\dfrac{S}{3\\pi}} = \\sqrt{\\dfrac{27}{3\\pi}} = \\dfrac{3}{\\sqrt{\\pi}}\\,\\mathrm{cm}."
For greater R "1.5\\pi R^2" will be greater than 0.5S, so the derivative will be negative, and for smaller values the derivative will be positive. So, "\\dfrac{3}{\\sqrt{\\pi}}\\,\\mathrm{cm}" is the point of maximum.
The corresponding volume is "0.5 \\left(27\\cdot\\dfrac{3}{\\sqrt{\\pi}} - \\pi \\cdot \\dfrac{27}{\\sqrt{\\pi}^3}\\right) = \\dfrac{27}{\\sqrt{\\pi}}" .
The height will be "h = \\dfrac{S-\\pi R^2}{2\\pi R} = \\dfrac{27-\\pi\\frac{9}{\\pi}}{2\\pi \\frac{3}{\\sqrt{\\pi}}} = \\dfrac{3}{\\sqrt{\\pi}}."
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