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Question 1027206: a wire 10 cm long is cut into two pieces, one of length x, while the other is of length 10-x. each piece is bent into a square.
a. find a function that models the total area enclosed by the two squares.
b. find the value of x that gives the minimum total area enclosed by the two squares.
c. find the minimum total area.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the area of a square = side * side = side squared.
s = side, a = area, formula is a = s^2.
you cut the wire into 2 pieces.
one piece has a length of x.
the other piece has a length of 10-x.
these lengths form the perimeter of the square.
there are 4 sides to the perimeter of a square.
x = perimeter, s = side, formula is x = 4s.
solve for s to get s = x/4.
the side of the first square is equal to x/4.
the side of the second square is equal to (10-x)/4.
the area of each square is equal to s^2.
the area of the first square is equal to (x/4)^2.
the area of the second square is equal to ((10-x)/4)^2
the total area of the two squares is equal to (x/4)^2 + ((10-x)/4)^2.
simplify each term in this expression to get:
(x/4)^2 = x^2/16.
((10-x)/4)^2 = (10-x)^2/16 = (100- 20x + x^2)/16
the expression become:
x^2/16 + (100 - 20x + x^2/16)
factor out the common term of 1/16 to get:
(1/16) * (x^2 + 100 - 20x + x^2)
combine like terms and reorder the terms in descending order of degree to get:
(1/16) * (2x^2 - 20x + 100)
factor out the common term of 2 to get:
(2/16) * (x^2 - 10x + 50)
this expression represents the area of the two squares.
let y be equal to the area of the 2 squares, and this expression becomes the equation of:
y = (1/8) * (x^2 - 10x + 50
this is a quadratic equation that you can factor by setting it equal to 0.
you get:
(1/8) * (x^2 - 10x + 50) = 0
you can solve for the roots of this equation, but you don't have to.
you want to find the minimum point of the quadratic.
the minimum point of the quadratic is found by using the formula of x = -b/2a.
in this equation, .....
a = 1
b = -10
c = 50
x = -b/2a = 10/2 = 5.
the quadratic equation is at a minimum when x = 5.
the value of y when x is equal to 5 is found by replacing x in the original equation.
that original equation is equal to y = (1/8) * (x^2 - 10x + 50) after we factored out the gcf.
the value of y is therefore equal to (1/8) * (5^2 - 10*5 + 50) which is equal to (1/8) * (25 - 50 + 50) which is equal to (1/8) * 25 which is equal to 3.125.
the minimum point of the quadratic equation is (5,3.125).
the value of y is 3.125 when the value of x is 5.
the minimum area is equal to 3.125 square centimeters when the value of x is equal to 5 centimeters.
you cut a pice 5 centimeters long and the remaining piece is (10-5) = 5 centimeters long.
those 2 pieces form the 2 squares where the sum of the areas is smallest.
the graph of the equation of the area of the two squares is shown below.
the first equation shown is the original equation.
the second equation shown is the simplified version of the original equation.
both those equations form the same graph, as they should, because they are equivalent equations.
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