Optimizing a rectangular garden with a fixed perimeter
A homeowner wants to enclose a rectangular garden using 40 meters of fencing. One side of the garden will be against a wall and does not need fencing. The homeowner wants to maximize the area of the garden. Additionally, they want to know the dimensions that will give an area of at least 150 square meters.
1. Determine the dimensions that maximize the area, and find the maximum area.
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Let the length of the garden parallel to the wall be ( x ) meters, and the width perpendicular to the wall be ( y ) meters. Since one side is against the wall, the fencing is used for three sides: ( x + 2y = 40 ). The area is ( A = xy ). Substitute ( x = 40 - 2y ): ( A = (40 - 2y)y = 40y - 2y^2 ). This is a quadratic function opening downward, so the maximum occurs at the vertex. The vertex is at ( y = -\frac{b}{2a} = -\frac{40}{2(-2)} = 10 ). Then ( x = 40 - 2(10) = 20 ). The maximum area is ( A = 20 \times 10 = 200 ) square meters. So the dimensions are 20 m by 10 m, with the 20 m side parallel to the wall.
2. Find the range of dimensions that will give an area of at least 150 square meters.
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We need ( A = 40y - 2y^2 \geq 150 ). Rearrange: ( -2y^2 + 40y - 150 \geq 0 ), or ( 2y^2 - 40y + 150 \leq 0 ). Divide by 2: ( y^2 - 20y + 75 \leq 0 ). Solve the quadratic equation ( y^2 - 20y + 75 = 0 ): discriminant ( = 400 - 300 = 100 ), so ( y = \frac{20 \pm 10}{2} ), giving ( y = 5 ) or ( y = 15 ). The inequality holds for ( 5 \leq y \leq 15 ). Then ( x = 40 - 2y ) ranges from ( x = 40 - 10 = 30 ) (when ( y=5 )) to ( x = 40 - 30 = 10 ) (when ( y=15 )). So the dimensions that give an area of at least 150 m² are: width ( y ) between 5 m and 15 m, and length ( x ) between 10 m and 30 m, with ( x + 2y = 40 ).