Question 626330
Suppose that A, B, and C are positive constants and that x + y = C. Show that the minimum value of Ax² + By²  occurs when  x = {{{BC/(A+B)}}} and y = {{{AC/(A+B) }}}
<pre>
We know that if A > 0, the minimum value of y = Ax² + Bx + C 
occurs when x = {{{-B/(2A)}}}

To avoid conflict of letters we re-write that as

We know that if P > 0, the minimum value of y = Px² + Qx + R 
occurs when x = {{{-Q/(2P)}}}

Since x + y = C, y = C - x, so

Minimum value of Ax² + By² = 

Minimum value of Ax² + B(C - x)² =

Minimum value of Ax² + B(C - x)(C - x) =

Minimum value of Ax² + B(C² - 2Cx + x²) =

Minimum value of Ax² + BC² - 2BCx + Bx² =

Minimum value of Ax² + Bx² - 2BCx + BC² =

Minimum value of (A + B)x² - 2BCx + BC² 

And by the rule above:

We know that if P > 0, the minimum value of y = Px² + Qx + R
 
occurs when x = {{{-Q/(2P)}}}

Let P = (A + B),  Q = -2BC, and  R = BC²

Minimum value of (A + B)x² - 2BCx + BC² 

occurs when x = {{{-Q/(2P)}}} = {{{-(-2BC)/(2(A+B))}}} = {{{BC/(A+B)}}}

and since y = C - x, the value of y when x = {{{BC/(A+B)}}} is

y = C - {{{BC/(A+B)}}} =  {{{C/1}}} - {{{BC/(A+B)}}} =  

{{{C/1}}}·{{{(A+B)/(A+B)}}} - {{{BC/(A+B)}}} = {{{C(A+B)/(A+B)}}} - {{{BC/(A+B)}}} =

{{{(AC+BC)/(A+B)}}} - {{{BC/(A+B)}}} = {{{(AC+BC-BC)/(A+B)}}} = {{{AC/(A+B)}}}.


Edwin</pre>