Hello Lila Bird,
I like to work such problems in general terms, which allows us to derive a formula that we can use for similar cases, and also to see how the various parameters affect the solution. So let's define:
$$A_P$$ = the area of the pool itself.
$$w_x$$ = the width of the walkway at the deep/shallow ends of the pool.
$$w_y$$ = width of the walkway along the sides of the pool.
$$A$$ = the area of the rectangular plot of land containing the pool and the surrounding walkway.
$$x$$ = horizontal length of plot.
$$y$$ = vertical length of plot.
Please refer to the following diagram:
View attachment 993
Thus, we may express the area of the plot as:
$$A(x,y)=xy$$
where we are constrained by:
$$A_P=\left(x-2w_x \right)\left(y-2w_y \right)\,\therefore\,y=\frac{A_P}{x-2w_x}+2w_y$$
And so we obtain the area of the plot in one variable $x$:
$$A(x)=x\left(\frac{A_P}{x-2w_x}+2w_y \right)$$
So, next we want to equate the first derivative to zero to find the critical value(s):
$$A'(x)=x\left(-\frac{A_P}{\left(x-2w_x \right)^2} \right)+(1)\left(\frac{A_P}{x-2w_x}+2w_y \right)=\frac{2\left(w_y\left(x-2w_x \right)^2-w_xA_P \right)}{\left(x-2w_x \right)^2}=0$$
Hence, this implies:
$$w_y\left(x-2w_x \right)^2-w_xA_P=0$$
Solving for $x$, and taking the positive root, we find the critical value:
$$x=\sqrt{\frac{w_x}{w_y}A_P}+2w_x$$
To determine the nature of the extremum associated with this critical value, we may use the second derivative test. We find:
$$A''(x)=\frac{4w_xA_P}{\left(x-2w_x \right)^3}$$
We can easily see that:
$$A''\left(\sqrt{\frac{w_x}{w_y}A_P}+2w_x \right)>0$$
Hence, the extremum is a minimum. Next we can find $y$ as follows:
$$y=\frac{A_P}{\sqrt{\frac{w_x}{w_y}A_P}}+2w_y=\sqrt{\frac{w_y}{w_x}A_P}+2w_y$$
Thus, we find the dimensions minimizing the plot of land subject to the constraint on the area of the pool are:
$$(x,y)=\left(\sqrt{\frac{w_x}{w_y}A_P}+2w_x, \sqrt{\frac{w_y}{w_x}A_P}+2w_y \right)$$
Now, to answer the specific problem given, we may plug in the data (in yards):
$$w_x=3,\,w_y=2,\,A_P=54$$
and we find:
$$(x,y)=(15,10)$$
