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Answer:
Wire should be cut in two parts with the length = 15.69 ft and 12.31 ft
Step-by-step explanation:
Length of the wire = 28 ft has been cut in two pieces.
One piece is used to form a square and remaining piece to form a circle.
Let the length of the wire which forms the square is 'l' ft.
Area of the square = (side)²
Perimeter of the square = 4(side) = l
Length of one side = [tex]\frac{l}{4}[/tex]
So, the area of the square = [tex]\frac{l^{2}}{16}[/tex] ft²
Now length of the remaining part = perimeter of the circle = (28 - l) ft
2πr = (28 - l)
r = [tex]\frac{28-l}{2\pi }[/tex]
Area of the circle formed = πr²
= [tex]\frac{\pi(28-l)^{2} }{4(\pi )^{2} }[/tex]
= [tex]\frac{(28-l)^{2}}{4\pi }[/tex]
Combined area of the square and circle
= [tex]\frac{l^{2}}{16}[/tex] + [tex]\frac{(28-l)^{2}}{4\pi }[/tex]
= [tex]\frac{l^{2}}{16}[/tex] + [tex]\frac{(28-l)^{2}}{4\pi }[/tex]
Now to maximize the area we will find the derivative of the area with respect to l.
[tex]\frac{dA}{dl}=\frac{d}{dl}[\frac{l^{2}}{16}+\frac{(28-l)^{2}}{4\pi}][/tex]
= [tex]\frac{d}{dl}[\frac{l^{2}}{16}+\frac{(28)^{2}+l^{2}-56l }{4\pi }][/tex]
= [tex]\frac{2l}{16}+\frac{(2l-56)}{4\pi }[/tex]
Now equate the derivative to zero.
[tex]\frac{2l}{16}+\frac{(2l-56)}{4\pi }[/tex] = 0
[tex](2l-56)=-\frac{2l}{16}\times 4\pi[/tex]
[tex](2l-56)=-\frac{\pi l}{2}[/tex]
[tex]2l+\frac{\pi l}{2}=56[/tex]
[tex]\frac{(4l+l\pi )}{2}=56[/tex]
l(4 + π) = 112
l(4 + 3.14) = 112
l = [tex]\frac{112}{7.14}[/tex]
l = 15.69 ft
Length of the other part = 28 - 15.69 = 12.31 ft
Therefore, wire should be cut in two parts with the length = 15.69 ft and 12.31 ft
The combined area formed by the circle and square will give U-shaped
parabolic curve, and therefore, will only have a minimum point, not a
maximum point in between
So that the combined area will be a minimum, the wire should be cut into
15.68 ft. and 12.32 ft. length pieces
Reason:
Let X represent the length of the piece used for the square;
The length of the piece used for the circle is therefore 28 - X
The area of the square, A₁ = [tex]\left(\dfrac{X}{4} \right)^2[/tex]
Circumference of the circle, C = 28 - X = 2·π·r
The radius of the circle is therefore;
[tex]r = \dfrac{28 - X}{2 \cdot r}[/tex]
The area of a circle, A₂ = π·r²
Therefore;
[tex]A_2 = \pi \cdot \left(\dfrac{28 - X}{2 \cdot \pi} \right)^2 = \dfrac{\left(28 - X\right)^2}{4 \cdot \pi}[/tex]
The area of the circle, A₂ = [tex]\dfrac{(28 - X)^2}{4 \cdot \pi}[/tex]
The sum of the area is therefore;
[tex]A = \left(\dfrac{X}{4} \right)^2 + \dfrac{(28 - X)^2}{4 \cdot \pi}[/tex]
[tex]A = \dfrac{(\pi + 4)\cdot X^2 + 224 \cdot X+ 3136}{16 \cdot \pi}[/tex]
From the above quadratic equation of the combined the coefficient of X²,
therefore, the graph has only a minimum point if the wire is used for both a
square and a circle.
At the minimum point, we have;
[tex]x = -\dfrac{b}{2 \cdot a}[/tex]
Therefore;
[tex]X = - \dfrac{\dfrac{224}{16 \cdot \pi} }{2 \times \dfrac{\pi + 4}{16 \cdot \pi} } = \dfrac{112}{\pi + 4} \approx 15.68[/tex]
The length of the wire to be used as a square to get the minimum area, X ≈
15.68 ft.
The length used for the circle is therefore; 28 ft. - 15.68 ft. ≈ 12.32 ft.
Therefore, the wire should be cut into 15.68 ft. and 12.32 ft. pieces.
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