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Maximizing garden area with fixed total fencing

A gardener wants to enclose a rectangular garden with dimensions xx by yy (in feet) and divide it into three equal-sized rectangular plots using two internal fences parallel to the sides of length xx. The total length of all fencing consists of the perimeter of the rectangle plus the two internal dividers. The total length of all fencing is fixed at 300 ft.

Let xx be the length (in feet) of the sides perpendicular to the internal fences, and let yy be the length (in feet) of the sides parallel to the internal fences. Two internal fences of length yy divide the rectangle into three equal plots. The total fencing used consists of the perimeter (two sides of length xx and two sides of length yy) plus two internal fences of length yy each. The total fencing satisfies 2x+2y+2y=300.2x + 2y + 2y = 300.

Using this constraint, express the total enclosed area AA as a function of xx alone, determine the domain of xx that makes physical sense, and find the dimensions xx and yy that maximize the enclosed area. What is the maximum possible area, in square feet?

Hint

Start by simplifying the fencing constraint to solve for yy in terms of xx, then substitute into A=xyA = xy. Remember that both xx and yy must be positive, which restricts the domain of xx. Find the critical point of A(x)A(x) and confirm it gives a maximum.

Step-by-step solution

The constraint is 2x+4y=3002x + 4y = 300, so solving for yy gives y=75−0.5xy = 75 - 0.5x. Since both x>0x>0 and y>0y>0 are required, we need 0<x<1500 < x < 150. The area function is then A(x)=xy=x(75−0.5x)=75x−0.5x2A(x) = xy = x(75 - 0.5x) = 75x - 0.5x^2.

Differentiating, A′(x)=75−xA'(x) = 75 - x, which is zero when x=75x = 75. Since A′′(x)=−1<0A''(x) = -1 < 0, this critical point is a maximum. Substituting back, y=75−0.5(75)=37.5y = 75 - 0.5(75) = 37.5, and the maximum area is A=(75)(37.5)=2812.5A = (75)(37.5) = 2812.5 square feet.

Maximizing garden area with fixed total fencing — solution graph

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