A piece of wire 26 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?

For the maximize the total area, the all wire i.e. 26 m should be used for the square.

Step-by-step explanation:

Length of the wire = 26 m

Let Amount of wire cut for square = x

Amount of wire cut for triangle  = 26 - x

Side of the square =

Area of the square =  ------ (1)

Side of the triangle is given by

Side of the triangle is

Area of the triangle is given by

Area of the triangle is

 ------- (2)

Now the total area = Area of square + Area of triangle

The total area =  ------- (3)

Differentiate above equation with respect to x we get

Take

------- (4)

By solving the above equation we get

x = 11.31 m

Again take by differentiating equation  (4)

Which is greater than zero. so the value x = 11.31 m gives the area minimum.

Thus for the maximize the total area, the all wire i.e. 26 m should be used for the square.

your answer is a

step-by-step explanation:

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the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation: