An arched window (an upper semi-circle and lower rectangle) has a total perimeter of 10 \ m. What is the maximum area of the window?

1 Answer
Nov 16, 2017

The maximum area is 50/(4 + pi) (~~ 7.001). This occurs when:

radius of the semicircle is 10/(4 + pi)
rectangular window is 10/(4+pi) xx 10/(4 + pi) (~~ 1.400 xx 1.400)

Explanation:

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Let us setup the following variables:

{ (r, "Radius of the semicircle","(m)"), (h=L, "Height of the rectangular window","(m)"), (A, "Total area enclosed by the window", "(sq m)") :}

Our aim is to find A(h,r), as a function of a single variable and to maximize the total area, A, wrt that variable (It won't matter which variable we do this with as we will get the same result). ie we want a critical point of A wrt the variable.

The total perimeter is that of 3 sides of the rectangle and the semicircle; we are told that this perimeter is 10 m

10 = (h + 2r + h) + (1/2)(2pir)
\ \ \ = 2r + 2h + pi r

:. 2h = 10 - 2r - pi r
:. \ \ h = 1/2(10 - 2r - pi r)

And the total Area is that of a rectangle and a semicircle:

A = (h)(2r) + (1/2)(pir^2)
\ \ \ = 2hr + 1/2 pi r^2
\ \ \ = 2(1/2(10 - 2r - pi r))r + 1/2 pi r^2
\ \ \ = 10r - 2r^2 - pi r^2 + 1/2 pi r^2
\ \ \ = 10r - 2r^2 - 1/2 pi r^2

We now have the Area, A, as a function of a single variable r, so differentiating wrt x we get:

(dA)/(dr) = 10 - 4r - pi r

At a critical point we have (dA)/(dr) =0 =>

10 - 4r - pi r = 0
:. \ \ \ \ 4r + pi r = 10
:. \ \ \ r(4 + pi) = 10
:. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \r = 10/(4 + pi) (~~ 1.400)

With this value of r we have:

A = 10(10/(4 + pi)) - 2(10/(4 + pi))^2 - 1/2 pi (10/(4 + pi))^2
\ \ \ = 50/(4 + pi) (~~ 7.001)

And:

h = 1/2(10 - 2(10/(4 + pi)) - pi (10/(4 + pi)))
\ \ = 10/(4+pi) (~~ 1.400)

We can visually verify that this corresponds to a maximum by looking at the graph of y=A(r):

graph{10x - 2x^2 - 1/2 pi x^2 [-5, 10, -5, 22]}

And also check that the perimeter is correct:

P = 2r + 2h + pi r
\ \ = 2(10/(4 + pi)) + 2(10/(4 + pi)) + pi(10/(4 + pi))
\ \ = 10