Differentiation (short maxima qn)

A piece of wire of length x cm is bet to form a rectangle. Show that the area of rectangle is maximum when it is a square.

Please show solution and explanation if possible thanks

If a rectangle has total perimeter x, and one side is z, then the other side is x/2 - z.

Thus the area is z(x/2 - z). = zx/2 - z^2.

Differentiate with respect to z:

A'(z) = x/2 - 2z

Set the derivative to 0 to find the maximum:

0 = x/2 - 2z

z = x/4

When z = x/4, the other side obviously equals x/2 - z = x/4 as well. It's a square.

Let length of one side of the rectangle be l. Area function A(l) is given by
A(l) = l*{(x/2)-l} = (x*l)/2 - l^2
dA/dl = x/2 - 2l = 0 gives l = x/4
d^2A/dl^2 = -2 which is negative always, so at l = x/4, Area is maximum
But at l = x/4, The rectangle is a square because other side is also (x/2) - x/4 = x/4. Hence proved.

If a and b are lengths the 2(a+b) = x or b = (x/2) - a
so area ab = a[(x/2)-a] = (ax/2) - a^2 = (x^2/16) - [ a-(x/4)]^2 < (x^2)/16 but this happens only if a = x/4 in which case b = x/4 and the rectangle is a square.

If a and b are sides , 2a+2b= x and A= ab
If A= A(a) , we do , dA/da=0 , we see that b= (x-2a)/2

A= a ( x-2a)/2 , x is a constant

dA/da= x/2 - 2a
2A/da2 = -2<0 thus, A will be Amax

dA/da=0 = x/2 -2a=0 or a= x/4 , thus b= (x-2a)/2
b= (x-x/2 )/2 , ie b= x/4 =a ok /

... x = 2L + 2W

... A = L * W
or A = L * ½ ( x - 2L )
or A = ½ x L - L^2
or dA/dL = ½ x - 2L = 0 at max/min
or L = ¼ x → W = ½ (x - 2(¼ x) ) = ¼ x

{ L, W } = { ¼ x, ¼ x } ← a square