Questions: A rectangular garden is to be built with 300, so that it is enclosed: - on three sides by a fence costing 10 per meter and - on one side by a fence costing 5 per meter. Find the dimensions of the garden that maximize the area of the garden. Make sure to use units in your final answer. Use a derivative test to justify your answer.

Solution Steps

We start by defining the variables for the dimensions of the garden. Let \( x \) be the length of the side with the cheaper fence (\$5 per meter) and \( y \) be the length of the other sides (\$10 per meter). The total cost for the fencing can be expressed as: \[ 5x + 10(2y + x) = 300 \] This simplifies to: \[ 15x + 20y = 300 \]

From the cost equation, we can solve for \( y \): \[ 20y = 300 - 15x \implies y = \frac{300 - 15x}{20} = 15 - \frac{3x}{4} \]

The area \( A \) of the garden can be expressed as a function of \( x \): \[ A(x) = x \cdot y = x \left(15 - \frac{3x}{4}\right) = 15x - \frac{3x^2}{4} \]

To find the critical points, we take the derivative of the area function with respect to \( x \): \[ A'(x) = 15 - \frac{3x}{2} \]

Setting the derivative equal to zero to find the critical points: \[ 15 - \frac{3x}{2} = 0 \implies \frac{3x}{2} = 15 \implies x = 10 \]

Next, we find the second derivative of the area function: \[ A''(x) = -\frac{3}{2} \] Since \( A''(x) < 0 \), this indicates that the critical point \( x = 10 \) corresponds to a maximum area.

Finally, we substitute \( x = 10 \) back into the equation for \( y \): \[ y = 15 - \frac{3(10)}{4} = 15 - \frac{30}{4} = 15 - 7.5 = \frac{15}{2} \]

Thus, the dimensions of the garden that maximize the area are \( x = 10 \) meters and \( y = \frac{15}{2} \) meters.

Final Answer