Question

Show that a rectangle with area of 100cm^2 and minimum perimeter is a square?

Answer 1

A square of side `10cm`

Explanation:

Let us set up the following variables:

# { (x, "length of the rectangle", cm), (y, "width of the rectangle", cm), (A=100, "Area of the rectangle", cm^2), (P, "Perimeter of the rectangle", cm) :} #

Our aim is to find `P(x,y)`, as a function of a single variable and to minimize the total area, `P`, wrt that variable (It won't matter which variable we do this with as we will get the same result). ie we want a critical point of `A` wrt the variable. Intuition tells us that a square is the optimum shape.

The total area of the rectangle is given by:

`A = xy`
`\ \ \ => xy = 100` ..... [A]

The total perimeter of the rectangle is given by:

`P = 2x + 2y`
`\ \ \ = 2x + 2(100/x)` from [A]
`\ \ \ = 2x + 200/x`

We now have the Perimeter, `P`, as a function of a single variable `x`, so differentiating wrt `x` we get:

`(dP)/(dx) = 2 - 200/x^2`

At a critical point we have `(dP)/(dx) =0 =>`

`2-200/x^2 = 0`
`:. (2x^2-200)/x^2 = 0`
`:. x^2 = 100`
`:. x = 10 \ \` (we obviously discard the -ve root)

With this value of `r` we have:

`y = 100/x = 10`

Thus `x=y=10 =>` A square

We can visually verify that this corresponds to a minimum by looking at the graph of `A(x)` rather than a maximum which we seek.

graph{2x+200/x [-50, 50, -84.4, 84.6]}