PRACTICE ANOTHER A piece of wire 18 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (a) How much wire should be used for the square in order to maximize the total area? m (b) How much wire should be used for the square in order to minimize the total area? m

Step-by-step explanation:

We have the equations

4x + 3y = 18   where x = the side of the square and y = the side of the triangle

For the areas:

A = x^2 + √3y/2* y/2

A = x^2  + √3y^2/4

From the first equation x = (18 - 3y)/4

So substituting in the area equation:

A = [ (18 - 3y)/4]^2 + √3y^2/4

A = (18 - 3y)^2 / 16 + √3y^2/4

Now for maximum / minimum area the derivative = 0 so we have

A' = 1/16 * 2(18 - 3y) * -3 + 1/4 * 2√3 y = 0

-3/8 (18 - 3y) + √3 y /2 = 0

-27/4 + 9y/8 + √3y /2 = 0

-54 + 9y + 4√3y = 0

y = 54 / 15.93

= 3.39 metres

So x = (18-3(3.39) / 4 = 1.96.

This is a minimum value for x.

So the total length of wire the square  for minimum  total area is 4 * 1.96

= 7.84 m

There is no maximum area as the equation for the total area is a quadratic with a positive leading coefficient.