Determining greatest and least values
The cuboid shown below has a volume of 72cm3.
Question
1. Show that the surface area of the cuboid is \(A(x) = 12\left( {{x^2} + \frac{{10}}{x}} \right)\)
1. \({A_{Front}} = length \times breadth\)
\(= 3x \times h\)
\(= 3xh\,unit{s^2}\)
\({A_{Back}} = length \times breadth\)
\(= 3x \times h\)
\(= 3xh\,unit{s^2}\)
\({A_{Left}} = length \times breadth\)
\(= 2x \times h\)
\(= 2xh\,unit{s^2}\)
\({A_{Right}} = length \times breadth\)
\(= 2x \times h\)
\(= 2xh\,unit{s^2}\)
\({A_{Top}} = length \times breadth\)
\(= 3x \times 2x\)
\(= 6{x^2}\,unit{s^2}\)
\({A_{Bottom}} = length \times breadth\)
\(= 3x \times 2x\)
\(= 6{x^2}\,unit{s^2}\)
Total Surace Area \(= 10xh + 12{x^2}\)
This still does not match what we require to prove as it has an \(h\) term, but we want an expression purely in terms of \(x\).
We can see from the beginning of the question that the volume of the cuboid is 72cm3.
\(Volum{e_{cuboid}} = length \times breadth \times height\)
\(= 3x \times 2x \times h\)
\(= 6{x^2}h\)
Therefore \(6{x^2}h = 72\)
\(h = \frac{{72}}{{6{x^2}}}\)
\(h = \frac{{12}}{{{x^2}}}\)
We can now substitute this into our expression for the total surface area.
\(A(x) = 10x \times \frac{{12}}{{{x^2}}} + 12{x^2}\)
\(A(x) = 12{x^2} + \frac{{120}}{x}\)
\(A(x) = 12 ( {x^2} + \frac{10}{x} )unit{s^2}\) as required.
Question
2. Calculate the minimum surface area
2. Again, in order to find the minimal surface area, we need to find which value of \(x\) gives a minimum turning point and then find its corresponding area.
\(A(x) = 12\left( {{x^2} + \frac{{10}}{x}} \right)\)
\(A(x) = 12{x^2} + \frac{{120}}{x}\)
\(A(x) = 12{x^2} + 120{x^{ - 1}}\)
\(A\textquotesingle (x) = 24x - 120{x^{ - 2}}\)
\(A\textquotesingle (x) = 24x - \frac{{120}}{{{x^2}}}\)
Stationary points occur when \(\frac{{dy}}{{dx}} = 0\).
\(24x - \frac{{120}}{{{x^2}}} = 0\)
Multiply each term by \({x^2}\)
\(24{x^3} - 120 = 0\)
\(24{x^3} = 120\)
\({x^3} = \frac{{120}}{{24}}\)
\({x^3}=5\)
\(x= \sqrt[3]{5}\)
Nature
\(\frac{{dy}}{{dx}} = 24(1) - \frac{{120}}{{{1^2}}} = - 96\) (negative)
\(\frac{{dy}}{{dx}} = 24(\sqrt[3]{5}) - \frac{{120}}{{{{(\sqrt[3]{5})}^2}}} = 0\) (stationary)
\(\frac{{dy}}{{dx}} = 24(2) - \frac{{120}}{{{2^2}}} = 18\) (positive)
Therefore minimum surface area when \(x = \sqrt[3]{5}\)
\(A(x) = 12\left( {{x^2} + \frac{{10}}{x}} \right)\)
\(A(\sqrt[3]{5}) = 12\left( {{{(\sqrt[3]{5})}^2} + \frac{{10}}{{\sqrt[3]{5}}}} \right)\)
\(A(\sqrt[3]{5}) = 105.3\,unit{s^2}(to1\,d.p.)\)
Therefore the minimum surface area is \(105.3\,unit{s^2}\)
