Paul wants to enclose part of his yard to make a rectangle dog run. He will put up a fence on the three sides of the rectangle and use the wall of the house as the fourth side of the rectangle. If he has 36 yd of fencing material, and wants the dog run to be the maximum are possible, what length should make the dog run? Assume that the length is longer than the width.We assume there is no fencing required along the red line, the side of the house. Let x be the width of the rectangle and L be the length of the rectangle. Then the Area, y, is given by (1) y = Lx Since we are told that Paul has 36 yd. of fencing, then x+L+x = 36 or 2x+L = 36 L = 36 - 2x We substitute that in equation (1): y = Lx y = (36 - 2x)x y = x(36 - 2x) y = 36x - 2x² y = -2x² + 36x This is the equation of a parabola that opens downward: Since the area is y, we need to find what point has the highest possible value for y. That will be at the highest point on the parabola. That highest point is the vertex. So we use the vertex formula: x-coordinate of vertex of parabola y = ax²+bx+c is -b/(2a) We compare y = -2x² + 36x to y = ax² + bx + c and find that a=-2, b=120 and c=0 so the x-coordinate of the vertex is -36/[2∙(-2)] = -36/(-4) = 9 So the maximum area will occur when x = 9, which will be the width of the rectangle, and since the length is given by L = 36 - 2x the length will be L = 36 - 2(9) = 36 - 18 = 18. Therefore the solution is to make the width be 9 yd. and the length be 18 yd. Edwin