Question 1125614
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Solve {{{ 4x+y=18}}} for y


{{{ 4x+y=18}}}


{{{ 4x+y-4x=18-4x}}} Subtract 4x from both sides


{{{ y=18-4x}}}


{{{ y=-4x+18}}}


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Now plug this into the function 


{{{z = x^2+y^2}}}


{{{z = x^2+(y)^2}}}


{{{z = x^2+(-4x+18)^2}}} Notice how y is replaced with -4x+18


{{{z = x^2+(-4x+18)(-4x+18)}}} Expand


{{{z = x^2+16x^2-144x+324}}} FOIL


{{{z = 17x^2-144x+324}}} Combine like terms


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The goal is to minimize z, so we want to find the smallest z value for some input x


The equation {{{z = 17x^2-144x+324}}} is in the form {{{z = ax^2+bx+c}}} with a = 17, b = -144, c = 324


The x coordinate of the vertex is {{{h = -b/(2a)}}}. Let's plug in the values for a and b.


{{{h = -b/(2a)}}}


{{{h = -(-144)/(2*17)}}}


{{{h = 144/34}}}


{{{h = 72/17}}}


This is the x value that makes z the smallest. So {{{x = 72/17}}} which leads to


{{{y = -4x+18}}}


{{{y = -4(72/17)+18}}}


{{{y = -288/17+18}}}


{{{y = -288/17+306/17}}}


{{{y = (-288+306)/17}}}


{{{y = 18/17}}}


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So the final answer is that {{{x = 72/17}}} and {{{y = 18/17}}}. This pair of values makes {{{z = x^2+y^2}}} the smallest possible, ie a minimum.
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