Bob wants to cut a wire that is 60 cm long into two pieces. Then he wants to make each piece into a square. Determine how the wire should be cut so that the total area of the two squares is a small as possible?

Dec 2, 2016

The 60 cm long wire should be cut so that you have 2 lengths of 30 cm each.

Explanation:

Let 1 of the cut pieces total length be $x$
Then the other piece is length $60 - x$

Let the area for square 1 be ${A}_{1}$
Let the area for square 2 be ${A}_{2}$
Let the sum of the areas be ${A}_{s}$

All sides of a square are of equal length so:

${A}_{1} = {\left(\frac{x}{4}\right)}^{2}$

${A}_{2} = {\left(\frac{60 - x}{4}\right)}^{2}$

${A}_{s} = {A}_{1} + {A}_{2} = {\left(\frac{x}{4}\right)}^{2} + {\left(\frac{60 - x}{4}\right)}^{2}$

${A}_{s} = {x}^{2} / 16 + \frac{{x}^{2} - 120 x + 3600}{16}$

${A}_{s} = \frac{2 {x}^{2}}{16} - \frac{120}{16} x + \frac{3600}{16}$

${A}_{s} = \frac{1}{8} {x}^{2} - \frac{15}{2} x + 225$
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This is a quadratic and as the ${x}^{2}$ term is positive it is of general shape of $\cup$

Thus the minimum area (${A}_{s}$) is at the vertex. Let me show you a trick to determining the vertex $x$ value.

Write as ${A}_{s} = \frac{1}{8} \left({x}^{2} - \frac{8 \times 15}{2} x\right) + 225$

${A}_{s} = \frac{1}{8} \left({x}^{2} - \textcolor{red}{\frac{{\cancel{8}}^{4} \times 15}{{\cancel{2}}^{1}}} x\right) + 225$

$\textcolor{g r e e n}{\text{The above is the beginning of the process to 'complete the square'}}$

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(\textcolor{red}{- 4 \times 15}\right) = + 30$

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So the 60 cm long wire should be cut so that you have 2 lengths of 30 cm each.