Area of stone path that students plan to cover is 80ft². x-width of a path xft*8ft+(xft*16ft-xft*xft)=80ft² 8x+16x-x²=80, -x²+24x-80=0 x²-24x+80=0, x₁,₂=(24⁺₋√(24²-4*80))/2 x₁,₂=(24⁺₋16)/2, x₁=16, x₂=4 a. Width of stone path is 4ft. b. I choose 4ft because it cannot be other solution which is 16ft. The length of hole rectangular area is 16 ft so the only logical solution of those two is 4ft.
(a) x = 4 First, let's calculate the area of the path as a function of x. You have two paths, one of them is 8 ft long by x ft wide, the other is 16 ft long by x ft wide. Let's express that as an equation to start with. A = 8x + 16x A = 24x But the two paths overlap, so the actual area covered will smaller. The area of overlap is a square that's x ft by x ft. And the above equation counts that area twice. So let's modify the equation by subtracting x^2. So: A = 24x - x^2 Now since we want to cover 80 square feet, let's set A to 80. 80 = 24x - x^2 Finally, let's make this into a regular quadratic equation and find the roots. 80 = 24x - x^2 0 = 24x - x^2 - 80 -x^2 + 24x - 80 = 0 Using the quadratic formula, you can easily determine the roots to be x = 4, or x = 20. Of those two possible solutions, only the x=4 value is reasonable for the desired objective. (b) There were 2 possible roots, being 4 and 20. Both of those values, when substituted into the formula 24x - x^2, return a value of 80. But the idea of a path being 20 feet wide is rather silly given the constraints of the plot of land being only 8 ft by 16 ft. So the width of the path has to be less than 8 ft (the length of the smallest dimension of the plot of land). Therefore the value of 4 is the most appropriate.
