At POTW Doggy Care, there is a need for a new outdoor dog run space. The layout of the dog run space is represented by \(PQRSTU\) in the diagram below.
The lengths of the two longer sides, \(QR\) and \(RS\), are to be the same, and the lengths of the two shorter sides, \(PQ\) and \(ST\), are to be the same. There will be right angles at each corner.
The dog run space is to be built using a fence along \(PQ\), \(QR\), \(RS\), and \(ST\), and using the walls of the daycare along \(PU\) and \(TU\). The total fencing to be used is \(30\text{ m}\). Determine the dimensions of the dog run space that will give the maximum area for the dog run.
Extend \(PU\) to \(RS\), letting the intersection point be \(V\). Then \(PV \perp RS\).
Let \(x\) represent the lengths, in metres, of both \(PQ\) and \(ST\). Let \(y\) represent the lengths, in metres, of both \(QR\) and \(RS\). Since \(PQRV\) is a rectangle, \(RV=PQ=x\) and \(VS=RS-RV=y-x\).
The total length of fencing from \(P\) to \(Q\) to \(R\) to \(S\) to \(T\) is \[PQ+QR+RS+ST=x+y+y+x=2x+2y\] Since the total amount of fencing used is \(30\text{ m}\), we have \(2x+2y=30\). Thus, \(x+y=15\) and \(y=15-x\). \[\begin{aligned} \text{Area of dog run}&=\text{Area }PQRV +\text{Area }VSTU\\ &=QR \times RV+VS \times ST\\ &=yx+(y-x)x\\ &= 2xy - x^2 \end{aligned}\] Substituting \(y=15-x\), this becomes \[\begin{aligned} \text{Area of dog run}&=2x(15-x)-x^2\\ &=30x-2x^2-x^2\\ &=-3x^2+30x \end{aligned}\] Completing the square, we have \[\begin{aligned} \text{Area of dog run}&=-3(x^2-10x)\\ &=-3(x^2-10x+5^2-5^2)\\ &=-3(x^2-10x+25)+75\\ &=-3(x-5)^2+75 \end{aligned}\]
This is the equation of a parabola which opens down from a vertex of \((5,75)\). Thus, the maximum area is \(75 \text{ m}^2\), and occurs when \(x=5\text{ m}\). When \(x=5\), we have \(y=15-x=15-5=10\text{ m}\).
Therefore, if \(QR = RS = 10\text{ m}\) and \(PQ = ST =5\text{ m}\), this gives a maximum area of \(75 \text{ m}^2\).