Answer to Question #220161 in Calculus for Siri Chandana

Question

A piece of wire 20 m long is cut into two pieces. One piece is bent into a square and the

other is bent into an equilateral triangle. How should the wire be cut so that the total area

enclosed is:

(a) maximum? (b) minimum?

Answer 1

Let x metres be the side of the square, so 4x metres are the total amount of wire used for the square (with $0 ≤ 4 x ≤ 20 ⇒ 0 ≤ x ≤ 5 )$ For the equilateral triangle 20−4x metres of wire remains, and the side will be $\frac{20 − 4 x}{ 3}$

Given the side l of an equilateral triangle, the height is, using Pitagora's theorem, $h = \frac{l}{ 2} \sqrt 3$ so the area of the triangle is:

$A = l ⋅ \frac{l}{ 2} \sqrt 3 ⋅ \frac1 2 = l^ 2 \frac{ {\sqrt3}}{4}$

$( A = \frac{b ⋅ h} 2 )$

So the total area is:

$A=x^2+(\frac{20-4x}{3})^2 \frac{\sqrt{3}}{4}\\ A'= 2x+ \frac{\sqrt{3}}{4}*\frac{1}{9}*20(20-4x)*(-4)\\ A'=2x-2\frac{\sqrt{3}}{9}(20-4x)\\ A' > 0 \space if\\ 2x-2\frac{\sqrt{3}}{9}(20-4x) > 0 \implies 18x-40\sqrt{3}+8\sqrt{3}>0 \\ x> \frac{20*3(3\sqrt{3}-4)}{81-48}\\ x> \frac{20}{11}(3\sqrt{3}-4)= \bar{x}\\$

So our function Area is decreasing in $[ 0 , \bar{x} )$ and growing in $( \bar{x}, 5]$ and so the point $( \bar{ x} , A ( \bar{x} ) )$ is a minimum of the function. The conclusion is that maximum (absolute maximum) is in one of the two sides of the interval of the definition of x, 0 or 5.

If  $x = 0$ the value of area will be: $A = 100 \frac{√ 3}{ 9} ≅ 19 . 2$

If $x = 5$  the value of the area will be: $A = 25$ , that is greater than the other value.

Therefore

a. Minimum = 19.2

b. Maximum = 25