A rectangular hole is to be cut in a wall for a vent. If the perimeter of the hole is 112 in. and the length of the diagonal is a minimum, what are the dimensions of the hole? Let the length and width of the hole be x and y, respectively, and let the length of the diagonal be D. Find an equation that describes the quantity to be minimized. Upper D squared equals x squared plus (56 minus x )squared (Type an equation.) The length of the diagonal is a minimum when the hole is 28 in. by 28 in.
Accepted Solution
A:
Answer: D = √ (2x² -112x + 3136Step-by-step explanation:From problem statement we have:Perimeter of the hole P = 112 inlength of the hole xwide of the hole yPerimeter is by definition P = 2*x + 2*y so y = [( P - 2* x)] ÷ 2y = ( P - 2*x ) ÷ 2 and P = 112 in y = ( 112 - 2*x) ÷ 2 ⇒ y = 56 - x (1)Triangle ABC (see attached file) is straight in A. Then diagonal D is:D² = x² + y² from equation (1) D² = x² + ( 56 - x )² Solving: D² = x² + 3136 + x² - 112*x ⇒ D² = 2*x² - 112*x + 3136Finally D = √ (2x² -112x + 3136