To maximize the area of two side-by-side rectangular dog runs with a total of 60 feet of fencing along three sides, let's solve the problem step by step.

Step 1: Define the variables

Let:

• $x$ = length of each rectangular run perpendicular to the garage.
• $y$ = width of the runs parallel to the garage.

Step 2: Total fencing constraint

The total fencing available is 60 feet. Since the garage provides one side (no fencing needed there), the fencing will be used for:

1. Two widths $y$ (one for each run).
2. Two lengths $x$ (one for each run).

Thus, the fencing constraint is: $2x + 2y = 60$ Simplify: $x + y = 30$ $y = 30 - x$

Step 3: Area of the dog runs

The total area $A$ of the two runs is given by: $A = x \cdot y$ Substitute $y = 30 - x$ into the area formula: $A = x(30 - x)$ $A = 30x - x^2$

Step 4: Maximize the area

To find the value of $x$ that maximizes the area, take the derivative of $A$ with respect to $x$ and set it to zero: $\frac{dA}{dx} = 30 - 2x$ Set $\frac{dA}{dx} = 0$: $30 - 2x = 0$ $x = 15$

Step 5: Find $y$

Using $y = 30 - x$: $y = 30 - 15 = 15$

Step 6: Verify the maximum

The second derivative of $A$ is: $\frac{d^2A}{dx^2} = -2$ Since the second derivative is negative, $A$ is maximized when $x = 15$.

Final Dimensions

The dimensions that maximize the area are:

• Length ($x$) = 15 feet
• Width ($y$) = 15 feet

The maximum area is: $A = 15 \cdot 15 = 225 \, \text{square feet.}$

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Questions to Expand Understanding:

1. How does the fencing constraint affect the dimensions of the runs?
2. Why is the second derivative test important in verifying the maximum area?
3. Can we apply this solution to non-rectangular areas? Why or why not?
4. What happens to the dimensions if the total available fencing changes?
5. How would the solution change if the runs were not side-by-side?

Tip:

Always verify constraints (like fencing length) by plugging the dimensions back into the original equation to ensure consistency.