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

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The discussion focuses on solving a calculus-based optimization problem involving a rectangular swimming pool with a specified area and surrounding walkway. The area of the plot is defined in terms of the pool's dimensions and walkway widths, leading to a formula for the total area. By applying calculus, critical values are determined to find the dimensions that minimize the plot area. The final dimensions for the rectangular plot, given the constraints, are calculated to be 15 yards by 10 yards. This solution effectively addresses the problem of minimizing land area while accommodating the pool and walkway specifications.
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Here is the question:

Calculus applied max/min problem?

Can someone please help me with this?

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.
 
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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:

$$(x,y)=(15,10)$$
 

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