SOLUTION: A wire 10 cm long is cut into two pieces, one of length x and the other of length x and the other of length 10-x. Each piece is bent into a shape of square. Which of the following

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Question 1113459: A wire 10 cm long is cut into two pieces, one of length x and the other of length x and the other of length 10-x. Each piece is bent into a shape of square. Which of the following is the model for the total area enclosed by the two squares of function of x?
a.
b.
c.
d.
After finding the total area, what is the value of x that minimizes the total area of the two squares? CHOICES(3 cm, 4cm, 5cm, and 6cm)
Found 2 solutions by math_helper, Theo:
Answer by math_helper(2461) (Show Source):
You can put this solution on YOUR website!
The first (x) length makes a square that has area
The 2nd length is 10-x, so it makes a square that has area

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I will use Calculus to find the minimum:
First we take the 1st derivative of A with respect to x (giving us the rate of change of A with respect to x):
�> This is zero when x = 5cm. So 5cm is a critical point.
The 2nd derivative of A with respect to x tells us if the function is curved up or curved down (at x=5):
so the curve is curved up (concave up) everywhere which means the critical point is at a minimum.
minimizes the area.
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Since this problem was multiple choice, the endpoints x=0 and x=10 need not be considered. In solving min/max problems in general, one must consider the function value at the endpoints in addition to at all critical points found.
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To figure out the minimum without Calculus, one way to see the minimum is to plot the graph of the function (x along the x-axis, A(x) along the y axis). Since this was multiple choice, you'd plot (compute A(x)) for the 4 values given, and compare their heights. When plotting a function in general you would take use x values and plot more A(x) values.

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
the wires are cut into lengths of x and 10-x.

they are then shaped into squares.

the wires form the perimeter of the squares.

the length of each side of the wire with length x, is equal to x/4.

the length of each side of the wire with length (10-x) is equal to (10-x) / 4

the area of each of the squares is equal to s^2, where s is the length of each side of the square.

the area of the square formed by the wire of length x would be s^2 = (x/4)^2.

the area of the square formed by the wire of length (10-x), would be s^2 = ((10-x)/4)^2.

the total area of both square is therefore (x/4)^2 + ((10-x)/4)^2

since (a/b)^2 is equal to a^2/b^2, the total area therefore becomes equal to x^2/4^2 + (10-x)^2/4^2

that simplifies to x^2/16 + (10-x)^2/16

that simplifies to (x^2 + (10-x)^2) / 16

(10 - x)^2 is equal to (10 - x) * (10 - x) which is equal to 10*10 -10x - 10x + x^2)

that simplifies to 100 - 20x + x^2.

therefore (x^2 + (10-x)^2) / 16 becomes (x^2 + 100 - 20x + x^2)/ 16.

that simplifies to (2x^2 - 20x + 100) / 16

that simplifies to (x^2 - 10x + 50) / 8

that looks like selection C.

you can confirm by assigning an arbitrary value to x that is within the limits of the problem.

the total length of the wire is 10.
if you assume that x = 8, then you have lengths of 8 and 2.
the wires are then formed into 2 squares.
the first square has a side length of 8/4 = 2
the second square has a side length of 2/4 = .5
the area of the first square is 2^2 and the area of the second square is .5^2.
2^2 + .5^2 = 4 + .25 which is equal to a total area of 4.25.

using the final formula of (x^2 - 10x + 50) / 8, and replacing x with 8, we get (8^2 - 10*8 + 50) / 8 becomes (64 - 80 + 50) / 8 which becomes 34/8 which becomes 4.25.

selection C looks good.