Question

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.

Answer 1

To find the dimensions of the rectangular hole with the minimum length of the diagonal, set up an equation based on the given information and find the derivative to solve for the minimum. The hole dimensions for the minimum length of the diagonal are 28 in by 28 in.

Step-by-step explanation:

To find the dimensions of the rectangular hole with the minimum length of the diagonal, we can set up an equation based on the given information. Let's assume the length of the hole is x and the width is y.

We know that the perimeter of the hole is 112 in, so we can write the equation: 2x + 2y = 112.

Additionally, we can use the Pythagorean theorem to relate the length of the diagonal D to the length and width of the hole: D^2 = x^2 + y^2.

To find the dimensions for the minimum length of the diagonal, we can find the derivative of the equation D^2 = x^2 + y^2 and set it equal to 0.

After solving the equation, we find that the hole is 28 in by 28 in.

Answer 2

D = √ (2x² -112x + 3136

Step-by-step explanation:

From problem statement we have:

Perimeter of the hole P = 112 in

length of the hole x

wide of the hole y

Perimeter is by definition P = 2*x + 2*y so y = [( P - 2* x)] ÷ 2

y = ( 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 + 3136

Finally D = √ (2x² -112x + 3136