Q:

# 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