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?

Answer 1

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)#

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 #

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Answer 2

To maximize the area of the window, the height of the rectangle should be half the height of the semi-circle, which is the radius of the semi-circle. Let the height of the rectangle be ( h ) and the radius of the semi-circle be ( r ). Since the total perimeter is 10 m, the perimeter equation is ( 2r + 2r + h = 10 ), simplifying to ( 4r + h = 10 ).

Now, the area ( A ) of the window is the sum of the area of the semi-circle and the rectangle, given by ( A = \frac{1}{2} \pi r^2 + r \cdot h ).

To express ( h ) in terms of ( r ) from the perimeter equation, we get ( h = 10 - 4r ). Substitute this into the area equation to get ( A = \frac{1}{2} \pi r^2 + r(10 - 4r) ).

Expand and simplify the area equation to get ( A = \frac{1}{2} \pi r^2 + 10r - 4r^2 ).

Differentiate ( A ) with respect to ( r ), set the derivative equal to zero to find critical points, and then check for maximum using the second derivative test. After simplifying, you'll find the maximum area occurs at ( r = \frac{5}{6} ) meters.

Substitute ( r = \frac{5}{6} ) into the area equation to find the maximum area, which is approximately ( 2.98 , \text{m}^2 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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