Question 1207563
<pre>
Let V = the value that we want to maximize or minimize

{{{V}}}{{{""=""}}}{{{x^2+y^2}}}, where {{{x+y}}}{{{""=""}}}{{{4}}} or {{{y=4-x}}}

{{{V}}}{{{""=""}}}{{{x^2+(4-x)^2}}}

{{{V}}}{{{""=""}}}{{{x^2+(16-8x+x^2)}}}

{{{V}}}{{{""=""}}}{{{2x^2-8x+16)}}}

That is a parabola opening upward, so it has a minimum value but no maximum
value.

Use the vertex formula to find the x-coordinate of the minimum point:

{{{-b/(2a)}}}{{{""=""}}}{{{-(-8)/(2*2)=2}}}

Substituting in 

{{{V[min]}}}{{{""=""}}}{{{2(2)^2-8(2)+16)}}}{{{8}}}

So the minimum value is 8, which occurs when x=2, and y = 4-x = 4-2 = 2.

Edwin</pre>