(OPTIMIZATION) A company has been contracted to build a storage shed. It takes the form of a rectangle that has been positioned within a semicircle.

Question

The base of the shed is 6 m long and the door must be at least 1 m vertically from the roof of the shed. 

 

1. What is the largest area the doorway can have and what are its dimensions? (give the era to the nearest 0.01 m^2 and the dimensions to the nearest cm)

 

 

THANKK YOU

Answer 1

To find the largest area for the doorway and its dimensions, we can follow these steps:

Let's denote the dimensions of the rectangle as follows:

- Width of the rectangle: x meters (this corresponds to the width of the doorway).

- Length of the rectangle: y meters.

We are given that the base of the shed is 6 meters long, which corresponds to the diameter of the semicircle. So, the radius of the semicircle is r = 6/2 = 3 meters.

The total area of the shed, A, is the sum of the area of the rectangle and half the area of the semicircle:

A = Area of rectangle + 0.5 * Area of semicircle

A = (xy) + 0.5 * π * (r^2)

Now, we need to express one of the variables (either x or y) in terms of the other to have a single-variable optimization problem. Since the door must be at least 1 meter vertically from the roof of the shed, we have:

y ≥ 1

Now, we need to express y in terms of x to substitute it into the area equation. We can use the fact that the height of the rectangle plus the radius of the semicircle equals y:

x + 1 = y

Now, we can rewrite the area equation in terms of x alone:

A(x) = (x * (x + 1)) + 0.5 * π * (3^2)

A(x) = (x^2 + x) + 4.5π

To find the largest area, we need to find the critical points of A(x) by taking its derivative with respect to x and setting it equal to zero:

A'(x) = 2x + 1 = 0

2x = -1

x = -1/2

However, this solution doesn't make sense in this context because the width of the doorway cannot be negative.

So, the critical point does not provide a maximum area. We must check the endpoints of the feasible interval (based on the constraint y ≥ 1):

1. x = 1

2. x = 6 (the width cannot exceed the base of the shed)

Now, calculate the corresponding y values for each x:

1. If x = 1, then y = x + 1 = 2.

2. If x = 6, then y = x + 1 = 7.

Now, we can calculate the areas for both cases:

1. A(1) = (1 * 2) + 0.5 * π * (3^2)

A(1) = 2 + 4.5π ≈ 15.71 m²

2. A(6) = (6 * 7) + 0.5 * π * (3^2)

A(6) = 42 + 4.5π ≈ 55.85 m²

The largest area for the doorway is approximately 55.85 square meters, and its dimensions are a width (x) of 6 meters and a height (y) of 7 meters.