The area of a rectangle is #A^2#. Show that the perimeter is a minimum when it is square?

Question

The area of a rectangle is #A^2#.  Show that the perimeter is a minimum when it is square?

Charles Anctil · Accepted Answer

See explanation below.

Consider a rectangle of sides #x# and #y#. The perimeter is: #p = 2(x+y)# so we want to minimize #p# subject to the constraint: #xy=A^2# or #xy-A^2 = 0# We then form the Lagrange function: #Lambda(x,y,lambda) = 2(x+y)+lambda(xy-A^2)# and equate the gradient of #Lambda# to zero: #(del Lambda) /(del x) =2+lambday=0# #(del Lambda) /(del y) =2+lambdax=0# #(del Lambda) /(del lambda) =xy -A^2=0# Combining the first two we have: #2+lambda y = 2 + lambda x# which needs to hold for any #lambda#, so it implies: #x = y# then from the third we have: #x = y = A# and #p=4A# This is certainly a stationary point for #p(x,y)#. To prove that it is a minimum we can proceed directly: suppose we have a rectangle with sides: #x= A+dx# #y = A^2/x = A^2/(A+dx)# Then: #p= 2(A+dx+A^2/(A+dx))= 2((A+dx)^2+A^2)/(A+dx) = 2(2A^2+2Adx+dx^2)/(A+dx) = (4A(A+2x)+2dx^2)/(A+dx) =4A +(2dx^2)/(A+dx) > 4A#

Isabella Ackerman · Answer

Let us set up the following variables: # {(x, "Width of the rectangle"), (y, "Height of the rectangle"), (P, "Perimeter of the rectangle") :} # Our aim is to find #P(x)#, (a function of a single variable) and to minimize P, wrt #x# (equally we could the same with #y# and we would get the same result).  ie we want a critical point of #(dP)/dx#. Now, the total areas is given as #A^2# (constant) and so: # \ \ \ A^2=xy #
# :. y=A^2/x# ..... [1] And the total Perimeter of the rectangle is given by: # P  = x+x+y+y #
# \ \ \ = 2x+2y # And substitution of the first result [1] gives us: # P = 2x+(2A^2)/x # ..... [2] We no have achieved the first task of getting the perimeter, #P#, as a function of a single variable, so Differentiating wrt #x# we get: # (dP)/dx = 2 - (2A^2)/x^2 # At a critical point we have #(dP)/dx=0 => # # 2 - (2A^2)/x^2 = 0#
# :.  \ \ \ A^2/x^2 = 1 #
# :.  \ \ \ \ \ x^2 = A^2 #
# :.  \ \ \ \ \ \ \x = +-A #
# :.  \ \ \ \ \ \ \x = A \ \ \ because x>0, " and " A>0# Hence we have #x=A => y=A# (from [1]), ie a square of side #A#  We should check that #x=A# results in a minimum perimeter.  Differentiating [2] wrt x we get: # (d^2P)/dx^2 = (4A^2)/x^3 > 0 " when " x=A# Confirming that we have a minimum perimeter QED

Aurora Alvarado · Answer

undefined To show that the perimeter of a rectangle with area A^2 is minimum when it is a square, we first express the area of the rectangle as A^2 = l * w, where l is the length and w is the width. Since the perimeter P of a rectangle is given by P = 2(l + w), we need to minimize this expression with respect to both l and w.

From the area expression, we can express one of the variables in terms of the other, such as l = A^2 / w. Substituting this into the perimeter equation, we get P = 2(A^2 / w + w).

To find the minimum of P, we take the derivative of P with respect to w, set it equal to zero, and solve for w. Then we plug this value of w back into the expression for P to find the corresponding value of l.

Taking the derivative and solving for w, we find w = A, and substituting this value back into the expression for l, we find l = A. Thus, the dimensions of the rectangle that minimize the perimeter are l = A and w = A, which means the rectangle is a square. Hence, the perimeter is minimum when the rectangle is a square.