Question

Use Lagrange multipliers to prove that the rectangle with maximum area that has a given perimeter \(p\) is a square.

Short answer

So, the rectangle with maximum area that has given perimeter \(p\) is a square.

Step-by-step solution

Method of Lagrange multipliers

To find the maximum and minimum values of\(f(x,y,z)\)subject to the constraint\(g(x,y,z) = k\)(assuming that these extreme values exist and\(\nabla g \ne {\bf{0}}\)on the surface\(g(x,y,z) = k)\):

(a) Find all values of\(x,y,z\), and\(\lambda \)such that\(\nabla f(x,y,z) = \lambda \nabla g(x,y,z)g(x,y,z) = k\)

And

(b) Evaluate\(f\)at all the points\((x,y,z)\)that result from step (a). The largest of these values is the maximum value of\(f\); the smallest is the minimum value of\(f\).

Assumption

Let\(A(x,y) = xy\)

Constraints\(g(x,y):2x + 2y = p\)

Using Lagrange multipliers, and solve for equations:

\(\nabla f = \lambda \nabla g\)

\(\begin{array}{c}\nabla (xy) = \lambda \nabla (2x + 2y)\\y + x = \lambda (2 + 2)\end{array}\)

Form equation

Compare both sides in above equation we have

\(\begin{array}{c}y = 2\lambda \,\,\,..(1)\\x = 2\lambda \,\,\,...(2)\\2x + 2y = p\,\,\,...(3)\end{array}\)

Find critical point

From equation (1) and (2) \(x = y\)

Using this in (3); \(x = y = \frac{p}{4}\)

The extreme point is \(\left( {\frac{p}{4},\frac{p}{4}} \right)\).

Now the physical nature of the problem says that there must be an absolute maximum area which has to occur at the extreme point of\({\rm{A}}(x,y)\)so it must occur at\(x = y = {\rm{p}}/4\)

And hence the rectangle with given perimeter will have maximum area if it is a square.