Review question

# If we know the area-sum and the perimeter-sum, what are the sides? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5505

## Solution

There are four separate square enclosures, three of side $\quantity{x}{yd.}$ and one of side $\quantity{y}{yd.}$ The sum of the areas of the enclosures is $\quantity{147}{sq. yd.}$ and the sum of their perimeters is $\quantity{96}{yd.}$ Obtain two equations containing $x$ and $y$ and solve them.

The area of a square of side $x$ is $x^2$, so the sum of the areas of the enclosures is $3x^2 + y^2$.

Hence, our first equation is $3x^2 + y^2 = 147$.

Now a square of side $x$ has perimeter $4x$, and so the sum of the perimeter of the enclosures is $(3 \times 4x) + 4y$.

Our second equation therefore is $12x + 4y = 96$.

In order to solve these, we’ll now use the second equation to express $y$ in terms of $x$. Rearranging, we get

\begin{align*} 3x + y &= 24 \\ \implies y &= 24 - 3x. \end{align*}

If we substitute this into the area equation above, we have

\begin{align*} 3x^2 + (24 - 3x)^2 &= 147 \\ \implies 3x^2 + 576 - 144x + 9x^2 &= 147 \\ \implies 12x^2 - 144x + 429 &= 0\\ \implies 4x^2-48x+143=0\\ \implies (2x-13)(2x-11)=0\\ \implies x = 5.5, 6.5. \end{align*}

Alternatively we could use the quadratic equation formula to solve here, to get the two solutions for $x$ given by

\begin{align*} x & = \frac{144 \pm \sqrt{144^2 - 4 \times 12 \times 429}}{24}\\ & = \frac{144 \pm \sqrt{144}}{24}\\ & = \frac{144 \pm 12}{24}. \end{align*}

So we again get $x_1 = 6.5$ and $x_2 = 5.5$.

Now by using the perimeter equation above, we find

\begin{align*} x_1 = 6.5, \quad \rightarrow \quad y_1 &= 24 - 3 \times 6.5 \\ &= 4.5 \\ x_2 = 5.5, \quad \rightarrow \quad y_2 &= 24 - 3 \times 5.5 \\ &= 7.5. \end{align*}

Thus there are two solutions, namely $x_1 = 6.5$ yd., $y_1 = 4.5$ yd. and $x_2 = 5.5$ yd., $y_2 = 7.5$ yd.

How ‘likely’ is this, that we find two viable solutions for $(x,y)$ here?

If we sketch the graphs of $3x^2 + y^2 = 147$ (an ellipse) and $3x + y = 24$ (a straight line), we are asking the line to cut the ellipse in two places, both in the first quadrant.

If the perimeter was $100$ yd. instead of $96$ yd. do we still get two solutions?