A piece of wire 25 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?

Wire used for square is 7.55 m.

Step-by-step explanation:

Let the length of wire cut for square be 'x'.

Given:

Total length of the wire = 25 m

Length of wire for square = 'x'

Length of wire for equilateral triangle =

Now, area of square (a) =

Area of equilateral triangle (e) is given as:

Now, total area (A) is given as:

Now, in order to maximize 'A', the derivative of area with respect to 'x' must be 0.

∴

Therefore, the length of the wire that is cut for square should be 7.55 m for maximum area.

y = - 6

step-by-step explanation:

given the point (- 2, y) lies on the graph then it is a solution of it's equation.

substitute x = - 2 into the equation for value of y

y = 3(- 2) = 3 × - 2 = - 6

the point on the graph is (- 2, 6)

the answer is a i belaeve atleast thats what my calculator just gfave me

step-by-step explanation: