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?
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.
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.