Question

two triangular pens are built against a barn. one hundred seventy meters of fencing are to be used for the three sides and the diagonal dividing fence (see figure). what dimensions maximize the area of the pen? the area of the pen is maximized if the side perpendicular to the barn is about \\(\square\\) meters long and the side parallel to the barn is about \\(\square\\) meters long. (round to two decimal places as needed.)

Explanation:

Step1: Define Variables

Let \( x \) be the length of the side perpendicular to the barn, and \( y \) be the length of the side parallel to the barn. The diagonal dividing fence is also \( x \) (since the two triangles are congruent, forming a rectangle with the barn and the parallel side). So the total fencing used is \( 3x + y = 170 \), so \( y = 170 - 3x \).

Step2: Area Formula

The area \( A \) of the pen (which is two congruent right triangles, so equivalent to a rectangle with length \( y \) and width \( x \)) is \( A = x \cdot y \). Substitute \( y \) from Step 1: \( A(x) = x(170 - 3x) = 170x - 3x^2 \).

Step3: Find Critical Points

Take the derivative of \( A(x) \): \( A'(x) = 170 - 6x \). Set \( A'(x) = 0 \): \( 170 - 6x = 0 \Rightarrow 6x = 170 \Rightarrow x = \frac{170}{6} \approx 28.33 \).

Step4: Find \( y \)

Substitute \( x \approx 28.33 \) into \( y = 170 - 3x \): \( y = 170 - 3(\frac{170}{6}) = 170 - \frac{170}{2} = 85.00 \).

Answer:

The side perpendicular to the barn is about \(\boldsymbol{28.33}\) meters long and the side parallel to the barn is about \(\boldsymbol{85.00}\) meters long.