Calculus question

Question

A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 46 feet?

Answer 1

Maximizing area for a given perimeter P (a general solution)

where r = the radius of the semicircle

Let x = 2r + 2L be the part of the perimeter that comes from the 3 sides of the rectangle.

P = πr + x

L = ½(x – 2r)

If you maximize the rectangular area (A₂), you maximize the whole area A = A₁ + A₂

Maximizing the rectangular area A₂

A₂ = 2rL = r(x – 2r) = rx – 2r²

dA₂/dt = x – 4r = 0 means that x = 4r

L = ½(x – 2r) = ½(4r – 2r) = ½(2r)

So area is maximized when L = r = ½W

P = πr + x = πr + 4r = (π + 4)r

which means that A₂ is maximized when

L = r = P/(π + 4)

W = 2r = 2P/(π + 4)

Maximum Total Area

A = A₁ + A₂ = ½πr² + 2rL = ½πr² + 2r² = ½(π + 4)r² = ½P²/(π + 4)

In the current problem P = 46 ft, so

L = r = 46/(π + 4) = 6.44 ft

W = 2(46)/(π + 4) = 92/(π + 4) = 12.88 ft

A = ½P²/(π + 4) = ½(46)²/(π + 4) = 148.1 ft²

Answer 2

The question asks for an area, so first derive the equation for the area of the entire window:

A = (Area of semicircle) + (Area of rectangle)

A = 0.5*π*r² + w*h

Since it is given that w = d = 2*r:

A = 0.5*π*r² + 2*r*h

The other information given concerns the perimeter, so derive that equation as well:

P = π*r + 2*h + w

46 = π*r + 2*h + 2*r

This equation can be used to solve for h:

2*h = 46 - π*r - 2*r

h = 23 - 0.5*π*r - r

Substitute this expression for h into the equation for area, and the resulting equation has only one independent variable, which will allow it to be minimized using the first derivative test:

A = 0.5*π*r² + 2*r*h

A = 0.5*π*r² + 2*r*(23 - 0.5*π*r - r)

A = 0.5*π*r² + 46*r - π*r² - 2*r²

A = (-0.5*π - 2)*r² + 46*r

A = 46*r - (0.5*π + 2)*r²

Take the first derivative to find critical points:

A' = 46 - 2*(0.5*π + 2)*r

A' = 46 - (π + 4)*r

0 = 46 - (π + 4)*r

r = 46 / (π + 4)

r ≈ 6.44114

The number line test can be used to check that this critical point is a local maximum, but it is sufficient to note that A(r) is a negative quadratic, which implies that there should be a single critical point that is the local (and absolute) maximum.

Finally, substitute this value for r back into A(r) to get the maximum area (a calculator may be helpful here):

A(6.44114) = 148.146 ft²

Answer 3

Let x = width of rectangular section and y = height of rectangular section.

Perimeter = x + 2y + (1/2)πx = 46 So, y = 23 - (1/2)x - (1/4)πx

Area = A = xy + (1/2)π(x²/4) = x(23 - (1/2)x - (1/4)πx) + πx²/8 = 23x - (1/2)x² - πx²/8

A' = 23 - x - πx/4 = 0

x(1 + π/4) = 23

x[(4 + π)/4] = 23 So, x = 92 / (4 + π) ≈ 12.9 ft and y ≈ 6.4 ft

Show that this value of x maximizes the area.