find the length and width of a rectangle whose perimeter is 44feet and whose area is 112 suare feet
Accepted Solution
A:
Answer:the dimensions are W = 8 ft and L = 14 ftStep-by-step explanation:Use the variables L and W.Then P = 44 ft = 2L + 2W, and A = 112 ft² = L·W.Reducing the first equation, we get 22 ft = L + W. Solving for L, we get L = 22 ft - W.Substituting 22 ft - W for L in the area equation, we obtain:A = 112 ft² = L · W = (22 ft - W) · W, or 112 ft² = (22 ft)W - W²Let's write this in standard form for a quadratic:112 ft² = (22 ft)W - W² ↔ W² - (22 ft) · W + 112 ft² = 0The coefficients of this quadratic equation are a = 1, b = -22 and c = 112.The discriminant is thus (-22)² - 4(1)(112), or 484 - 448 = 36.Thus, the roots are: -(-22) ± √36W = --------------------- 2 22 ± 6 = -------------- , so that W = 14 and W = 8. 2Since L = 22 - W, L could be either 22 - 8 = 14 or 22 - 14 = 8Thus, the dimensions are W = 8 and L = 14.Check: Does WL = 8(14) ft = 112 ft²? YES Does P = 2W + 2L = 2(8 ft) + 2(14 ft) = 44 ft? YES