Find the total cost of the glass and maximum area.

Question

A glass window is in the shape of a rectangle capped with a semi circle. The perimeter is not to exceed 30m. The price for glass is $9.75/m². Find the total cost of the glass needed and the maximum area.

Answer 1

Let x be the radius of the semi circle. let y be the height of the window, the width of the window would be 2x.

 

The perimeter of the window is 2x + 2y + pi*x = 30

The Area of the window is 2xy + .5pi x^2 which is what we want to maximize.

 

Solving the perimeter equation for y we get....y = 15 - x - pi/2 *x. Substitute in for y in the area expression and get

 

A(x) = 2x(15 - x - pi/2*x) + .5pi x^2

A(x) = 30x - 2x^2 - pi x^2 + .5pi x^2

A(x) = 30x - 2x^2 - .5pi x^2

 

Take the derivative.....I'm assuming this is a calculus problem. If not, you needto find the axis of symmetry.

A'(x) = 30 - 4x - pi x

 

Set it equal to 0 and solve,

30 - 4x - pi x = 0

30 = 4x + pi x

30 = (4 + pi)x

30/(4+pi) = x

x is approximately 4.2

 

Plug back in: 2(4.2) + 2y + pi*4.2 = 30

21.6 + 2y = 30

2y = 8.4

y = 4.2

 

Plug x = 4.2 into the area formula to get the maximum area.

A(4.2) = 30(4.2) - 2(4.2)^2 - .5pi(4.2)^2 =126 - 35.28 - 27.71 = 63 sq meters

 

63 sq meters at $9.75 per square meter would be $614.25.

 

I'm not sure if this is correct so I'm going to calculate A(4.1) and A(4.3), Both should lead to an area less than 63.

A(4.1) = 62.97 and A(4.3)= 62.98

 

I am now relatively sure that my answer is correct.

Answer 2

Lets look first at the geometry of the problem, then go from there. There's a semi-circle on the top of the window, so let's call its radius R. The width of the rectangular part of the window then has to be 2R. The height we can call H.

 

The perimeter of the window includes two heights and the base of the rectangle, and the circumference of the semicircle on top. A full circle has circumference C = 2πR, so a semicircle has C = πR. The total perimeter is then:

P = 2(H) + (2R) + πR

 

The area is a rectangle plus a semicircle:

Area = H*(2R) + 1/2 (πR²)

 

It's cost will be $9.75 * Area

 

Now let's look at the other information we have. The maximum perimeter is 30m, so we'll set P=30. With this, we can solve the perimeter equation for either H or R (I recommend H), which we can then substitute into the area equation in place of H.

 

You'll then have a messy quadratic equation for Area = (stuff with Rs and R²s). Collect the terms so that you get something in the form of Area = AR² + BR + C. (You'll probably want to plug in 3.14159 or use your calculator to find numerical values for A, B, and C rather than finding algebraic expressions for each.) The maximum or minimum value of a quadratic equation occurs at its vertex, which you can find (in this case) its R = -B/(2A).

 

Almost done! With this value of R, you can find H from the perimeter formula. With both H and R, you can find the area, and then the cost. Whew.