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Question: A farmer wants to construct a rectangular pigpen using 200 ft of fencing. The pen will be built…

A farmer wants to construct a rectangular pigpen using 200 ft of fencing. The pen will be built next to an existing stone wall, so only three sides of fencing need to be constructed to enclose the pen. What dimensions should the farmer use to construct the pen with the largest possible area?

Solution

To solve this problem, the goal is to maximize the area of a rectangular pigpen that uses 200 feet of fencing along three sides, with one side against an existing wall. Let the width of the pen perpendicular to the wall be \( x \). The length parallel to the wall would then be \( 200 - 2x \), because the farmer uses two widths and one length of fencing. The area \( A \) is given by the formula for the area of a rectangle: \[ A = x(200 - 2x) \] Expand the area equation: \[ A = 200x - 2x^2 \] This is a quadratic equation of the form: \[ A = -2x^2 + 200x \] To find the maximum area, we need the vertex of the parabola represented by this quadratic equation. The vertex form \( x \) for a quadratic equation \( ax^2 + bx + c \) is: \[ x = -\frac{b}{2a} \] Substitute \( a = -2 \) and \( b = 200 \): \[ x = -\frac{200}{2(-2)} = \frac{200}{4} = 50 \] The width \( x \) is 50 feet. Substitute \( x = 50 \) back into the expression for the length: \[ 200 - 2x = 200 - 2(50) = 200 - 100 = 100 \] Therefore, the dimensions that give the largest possible area are 50 feet for the width and 100 feet for the length.

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