Lila Bird's question at Yahoo! Answers regarding minimizing plot of land

MarkFL

Staff member
Here is the question:

Calculus applied max/min problem?

A rectangular swimming pool is to have a area of 54 sq yards the walkway that surrounds the pool is 3 yards wide at the deep and shallow ends and 2 yards wide along the sides. Find the dimensions of the rectangular plot of the smallest area that can be used
I have posted a link there to this topic so the OP can see my work.

MarkFL

Staff member
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:

$$\displaystyle A_P$$ = the area of the pool itself.

$$\displaystyle w_x$$ = the width of the walkway at the deep/shallow ends of the pool.

$$\displaystyle w_y$$ = width of the walkway along the sides of the pool.

$$\displaystyle A$$ = the area of the rectangular plot of land containing the pool and the surrounding walkway.

$$\displaystyle x$$ = horizontal length of plot.

$$\displaystyle y$$ = vertical length of plot.

Please refer to the following diagram: Thus, we may express the area of the plot as:

$$\displaystyle A(x,y)=xy$$

where we are constrained by:

$$\displaystyle A_P=\left(x-2w_x \right)\left(y-2w_y \right)\,\therefore\,y=\frac{A_P}{x-2w_x}+2w_y$$

And so we obtain the area of the plot in one variable $x$:

$$\displaystyle A(x)=x\left(\frac{A_P}{x-2w_x}+2w_y \right)$$

So, next we want to equate the first derivative to zero to find the critical value(s):

$$\displaystyle A'(x)=x\left(-\frac{A_P}{\left(x-2w_x \right)^2} \right)+(1)\left(\frac{A_P}{x-2w_x}+2w_y \right)=\frac{2\left(w_y\left(x-2w_x \right)^2-w_xA_P \right)}{\left(x-2w_x \right)^2}=0$$

Hence, this implies:

$$\displaystyle w_y\left(x-2w_x \right)^2-w_xA_P=0$$

Solving for $x$, and taking the positive root, we find the critical value:

$$\displaystyle x=\sqrt{\frac{w_x}{w_y}A_P}+2w_x$$

To determine the nature of the extremum associated with this critical value, we may use the second derivative test. We find:

$$\displaystyle A''(x)=\frac{4w_xA_P}{\left(x-2w_x \right)^3}$$

We can easily see that:

$$\displaystyle A''\left(\sqrt{\frac{w_x}{w_y}A_P}+2w_x \right)>0$$

Hence, the extremum is a minimum. Next we can find $y$ as follows:

$$\displaystyle y=\frac{A_P}{\sqrt{\frac{w_x}{w_y}A_P}}+2w_y=\sqrt{\frac{w_y}{w_x}A_P}+2w_y$$

Thus, we find the dimensions minimizing the plot of land subject to the constraint on the area of the pool are:

$$\displaystyle (x,y)=\left(\sqrt{\frac{w_x}{w_y}A_P}+2w_x, \sqrt{\frac{w_y}{w_x}A_P}+2w_y \right)$$

Now, to answer the specific problem given, we may plug in the data (in yards):

$$\displaystyle w_x=3,\,w_y=2,\,A_P=54$$

and we find:

$$\displaystyle (x,y)=(15,10)$$