Respuesta :
Answer:
A(max) = 72,13 ft²
Step-by-step explanation:
Total area of the Norman window is area of the semicircle plus area of the rectangle
Let call x radius of the semicircle, then the width of the rectangle is 2*x
Area of semicircle A(sc) = π*r²/2 ⇒ A(sc) = π*x²/2
Area of rectangle ( sides 2*x and y) A(r) = 2*x*y
Area of window A(w) = π*x²/2 + 2* x*y
Perimeter of the window is:
Three sides of the rectangle P(r) = 2*y + 2*x
Length of the semicircle P(sc) = π*x
Then 35 = 2*y + 2*x + π*x ⇒ 2*y = 35 - 2*x - π*x
y = 35 - x* ( 2 + π ) /2
Area of window as a function of x
A(x) = π*x²/2 +2* [ 35 -x ( 2 +π ) /2 ]*x
A(x) = π*x²/2 + x* (35 - 2*x - π*x )
A(x) = π*x²/2 + 35*x - 2*x² - π*x²
Taking derivatives on both sides of the equation
A´(x) = π*x + 35 - 4*x - 2* π*x
A´(x) = 0 -π*x - 4*x = - 35 ⇒ - x ( 4 + π ) = -35 ⇒ x = 4,90 ft
And y = ( 35 - x - π*x )/2 ⇒ y = (35 - 4,90 - 15,38 )/2
y = 7,36 ft
A(max) = 2*x*y * + π*x²/2
A(max) = 72,13 ft²
The largest possible Norman window with a perimeter of 35 feet will have area as: 85.54 sq. ft approx.
How to obtain the maximum value of a function?
To find the maximum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.
Putting those values of x in the second rate of function, if results in negative output, then at that point, there is maxima. If the output is positive then its minima and if its 0, then we will have to find the third derivative (if it exists) and so on.
For this case, we'll first find the area of the Norman window in terms of its length and width. Using the perimeter given will make us express that area in terms of single variable.
Let we've got:
- Length of the rectangle in the window = L feet
- Width of the rectangle in the window = W feet
Thus, perimeter of the window = 2L + W + perimeter of semicircle with diameter W feet.
Perimeter of the semicircle with diameter W feet is: [tex]\dfrac{2 \times \pi \times (d/2)}{2} = \dfrac{\pi d}{2} = \dfrac{\pi .W}{2} \: \rm ft[/tex]
Thus, perimeter of the window = [tex]2L + W + \dfrac{\pi W}{2}[/tex]
Since perimeter of the window is given to be 35 feet, thus, we get:
[tex]2L + W + \dfrac{\pi W}{2} = 35\\\\L = \dfrac{1}{2}(35 - W(1+\dfrac{\pi}{2})) \approx 17.5 -1.285W[/tex]
The area of the window = area of rectangle + area of semicircle
or
[tex]A_{window} = L.W + \dfrac{\pi (d/2)^2}{2} = L.W + \dfrac{\pi .W^2}{8} \: \rm ft^2[/tex]
Putting value of L in terms of W, we get that area of window as:
[tex]A_{window} = f(W) \approx (17.5-1.285W)W + 0.39W^2 = 17.5W - 0.895W^2 \: \rm ft^2[/tex]
Now, finding the maxima of f(W) will give us the maximum area of the considered Norman window with perimeter 35 feet.
Finding the first and second rate of f(W), we get:
[tex]f'(W) = 17.5 - 1.79W\\f''(W) = -1.79 < 0[/tex]
Since second derivative is negative, all critical values we obtain from f'(W) = 0 would be maxima.
Getting critical values:
[tex]f'(W) = 17.5 - 1.79W=0\\W \approx 9.78 \: \rm ft[/tex]
This is the only critical value, so maximum of the function f(W).
Thus, maximum area of the function is evaluated at W ≈ 9.78 ft as:
[tex]max(A) = f(9.78) \approx 17.5(9.78) - 0.895(9.78)^2 \\max(A) \approx 85.54 \: \rm ft^2[/tex]
Thus, the largest possible Norman window with a perimeter of 35 feet will have area as: 85.54 sq. ft approx.
Learn more about maxima and minima of a function here:
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