Question 168001
area of a square is s^2 where s is the length of each side.
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the wire is cut into 2 lengths and forms a square with each.
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the wire of x length forms 4 sides with the length of each side equal to x/4.
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the wire of (10-x) length forms 4 sides with the length of each side equal to (10-x)/4.
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area of the square made by the wire of x length is {{{(x/4)^2}}}
this equals {{{x^2 / 4^2}}} which equals {{{x^2/16}}}
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area of the square made by the wire of (10-x) length is {{{((10-x)/4)^2}}}.
this equals {{{(10-x)^2 / 4^2}}} which equals {{{(10-x)^2 / 16}}}.
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the total area enclosed by these squares is given by the equation:
{{{f(x) = x^2/16 + (10-x)^2/16}}}
since the denominator is equal, this equation becomes:
{{{f(x) = (x^2 + (10-x)^2)/16}}}
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since x^2 is positive, this is a quadratic equation that opens upward and points downward.
the point where the graph turns is a minimum point.
that point can be found as follows:
take the given formula and multiply it out to get a quadratic equation in standard form.
that standard form is:
{{{(2*x^2 - 20*x + 100)/16}}}
that becomes:
{{{(x^2/8) - (10*x/8) + (50/8)}}}
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in that equation, the a coefficient is 1/8, and the b coefficient is -10/8.
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the formula {{{-b/2a}}} gives the minimum/ maximum point of a quadratic equation.
substituting, that formula becomes:
{{{- (-10/8) / (2/8) = 5}}}
when x = 5, y = 3.125
that's the minimum value of the equation.
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graph of the resulting equation looks like this showing that when x = 5, y = 3.125 and the value of the equation of the squares is at a minimum.
{{{graph(800,800,-1,9,-10,10,(x^2 + (10-x)^2)/16)}}}