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I am trying to help my daughter with this question:

A piece of wire 12 cm long is cut into 2 pieces. One piece is used to form a square and the other a rectangular shape in which the length is twice its width.
a. If x cm is the side length of the square, write down the dimensions in terms of x.
b. Formulate a rule for A, the combined area of the square and the rectangle in cm^2, in terms of x.
c. Determine the lengths of the two pieces if the sum of the areas is to be a minimum.

I have been able to solve the first two parts of the question. I am stuck at solving the third. Can you please point me in the right direction, using quadratics?

What I have solved so far:

a. Total length of string used for rectangle = 12 - 4x, therefore, 
   width of rectangle = (12 - 4x)/6 and length = (12 - 4x)/3
b. Combined area = (17x^2 - 48x + 72)/9
Rosie Smith
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1 Answers1

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We cut the wire into lengths $x, 12 - x$

If the perimeter of the square is $x$ the side length is $\frac {x}{4}$ and the area $\frac {x^2}{16}$

The short side of the rectangle is $(2 - \frac {1}{6} x)$ and the long side is $(4 - \frac {1}{3}x)$ The area is $(2-\frac 16 x)(4-\frac 13 x) = 8 - \frac {4}{3} x + \frac {1}{18} x^2$

Adding the square and the rectangle:

$A = \frac {x^2}{16} + 8 - \frac {4}{3} x + \frac {1}{18} x^2 = \frac {17}{144} x^2 - \frac {4}{3} x + 8$

$A$ is minimized when $x$ lies on the axis of symmetry.

$x = -\frac {b}{2a}$

$x = \frac {4\cdot 72}{3\cdot 17} = \frac {12\cdot 8}{17}$

I didn't fully multiply that out, because it will make the cancelations in the multiplications that follow a little easier.

$A = (\frac {17}{144}) (\frac {12\cdot 8}{17})^2 - (\frac 43)(\frac {12\cdot 8}{17}) + 8\\ \frac {64}{17} - \frac {128}{17} + 8 \\ 8 - \frac {64}{17} \\ \frac {72}{17}$

user317176
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