Question

A piece of wire of length 60 centimetres is cut into two sections.Each section is then bent into the shape of a square.FInd the smallest possible value of the sum of the areas of the two squares.

Answer 1

The smallest possible value of the sum of the areas of the two squares is 1800 square centimeters.

A piece of wire of length $L$, in centimetres, is cut into two sections, whose lengths are $x$ and $L-x$, both in centimetres, respectively. The equation of sum of the areas of the two squares is represented below:

$A = (L-x)^{2} + x^{2}$

$A = L^{2}-2\cdot L\cdot x +2\cdot x^{2}$ (1)

Where $A$ is the sum of the areas of the two squares, in square centimeters.

Now we proceed to determine the absolute minimum by means of first and second derivative tests: ($L$ - Constant, $A, x$ - Variables)

First derivative test

$A' = -2\cdot L + 4\cdot x$

$-2\cdot L + 4\cdot x = 0$

$x = \frac{L}{2}$

Second derivative test

$A'' = 4$

Which means that the critical value represents an absolute minimum.

Lastly, we calculate the area for $L = 60\,cm$:

$x = \frac{60\,cm}{2}$

$x = 30\,cm$

$A = 60^{2}-2\cdot (60)\cdot (30)+2\cdot (30)^{2}$

$A = 1800\,cm^{2}$

The smallest possible value of the sum of the areas of the two squares is 1800 square centimeters.

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