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?
The maximum area is
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
The total perimeter is that of
# 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,
# (dA)/(dr) = 10 - 4r - pi r #
At a critical point we have With this value of And: We can visually verify that this corresponds to a maximum by looking at the graph of graph{10x - 2x^2 - 1/2 pi x^2 [-5, 10, -5, 22]} And also check that the perimeter is correct:
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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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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.
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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