A wire of length L can be used to make a circle and a square. How much of the wire should be used for each of the figures to maximise area?
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The total length of the wire is L. Let us use it to make a square of length x and a circle of radius r. The combined area of the shapes is x^2 + pi*r^2. The circumference of the square is 4x and that of the circle is 2*pi*r
4x + 2*pi*r = L
We have to maximize A = x^2 + pi*r^2
Differentiate L and A with respect to r
dA/ dr = 2*x(dx/dr) + 2*pi*r
dL / dr = 4(dx/ dr) + 2*pi
As L is a constant dL/dr = 0
=> 4(dx/ dr) + 2*pi = 0
=> dx / dr = -2*pi/4 = -pi/2
substitute in dA/dr
=> dA/ dr = 2*x(-pi/2) + 2*pi*r
=> dA/dr = -pi*x + 2*pi*r
Take the second derivative with respect to r
=> d^2A / dr^2 = 2*pi – pi*(dx/dr)
=> d^2A / dr^2 = 2*pi + pi^2/2
The second derivative is always positive. The function of A versus r is concave upwards.
In the interval 0 <= 2*pi*r <= L, the function A takes the maximum value at either of r = 0 or r = L or both.
At r = 0, x = L/4, we find A = L^2 / 16
At r = L/2*pi, x = 0, we have A = L^2/ 4*pi
This gives the maximum value of A at r = L/ 2*pi
Therefore we can conclude that for the maximum area the wire should be used only to make the circle and no part of it used for the square.
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WE will assume that we will use x of the wire for the square and (L-x) of the wire for the circle.
Now the area of the square is given by:
A= side^2 = (x/4)^2 = x^2 /16
Now the area of the circle is given by:
A = r^2 * pi
But we have the circumference C = L-x
==> 2*pi*r = L-x
==> r = (L-x)/2pi
==>A = (L-x)^2/4*pi^2 * pi = (L-x)^2/4pi
Let us find the maximum area of the circle.
==> A' = 2(L-x)*-1 / 4pi = (x-L)/2pi = 0
==> x = L ...........(1)
For the square:
A' = 2x/16 = x/8
The maximum values is 0.
Then, the maximum area is when the wire used to make a circle only.
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