A piece of wire 23 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. Required:
a. How much wire must be used for the square in order to maximize the total area?
b. How much wire must be used for the square in order to minimize the total area?

wire for square to maximize total area = 23 m

Wire to minimize total area = 2.019 m

Step-by-step explanation:

For the square, let's say the total length of the square is "y" m.

Thus, length of one side is = y/4

And area of the square = (y/4) = y²/16

Since the wire is 23 m, then total length of equilateral triangle is; 23 - y.

Thus, length of one side of equilateral triangle = (23 - y)/3

Using trigonometric ratio, we can find the height of the triangle and thus area.

Area of triangle = (√3)/4) × ((23 - y)/3)²

Area of triangle = (√3)/36)(23 - y)²

So, total area of square and triangle is;

A_total = (y²/16) + (√3)/36)(23 - y)²

Now, extremizing this function by derivatives, we have;

dA/dy = (y/8) - (√3)/18)(23 - y)

d²A/dy² = ⅛ + (√3)/18)

So, d²A/dy² > 0

Now,let's find the maximum or minimum of the function.

So, we equate dA/dy to zero.

Thus;

(y/8) - (√3)/18)(23 - y) = 0

y/8 = (√3)/18)(23 - y)

(y/8) + (√3)/18)y = 23((√3)/18)

Multiply through by 8 to give;

y + 0.0962y = 2.2132

1.0962y = 2.2132

y = 2.2132/1.0962

y = 2.019 m

2.019 will be a minimum because d²A/dy² > 0

The maximum will occur at a boundary of the allowed values. Thus, the absolute maximum is for y = 23.

Note that a square has more area than a triangle and as such it is normal for the square to get the largest area over the triangle and therefore we will have to use all of the wire to construct the square.

answer: evaluate the objective function at each vertex to determine which - and -values, if any, maximize or minimize the function. example: find the minimum value and maximum value of the objective function f ( x , y ) = 4 x + 5 y , subject to the following constraints

3030 ur welcome

step-by-step explanation: