6
$\begingroup$

How to find minimum of the expression $$\, \big(\!-x+y+1 \big)^2 + \big( x-y-2\big)^2 + \big(x+2y-3 \big)^2 \,$$ without using partial derivatives?

It is easy to find the answer $\; x = 2, \; y = \dfrac{1}{2}\; $ by computing gradient of the expression above, but I do not see the way to do so without using partial derivatives.

This is the part of homework for linear algebra class designed for freshmen. I feel that there is no way firs-year students are expected to use partial derivatives because this topic is only taught in the end of CALC II class. I feel very dumb and discouraged since I could not help the student. We tried to make substitution $\;z = x-y,\;$ or to expand the brackets, but nothing seemed to give definite answer. Any help is appreciated.

$\endgroup$
12
$\begingroup$

By completing the square, $$ \big(-x+y+1 \big)^2 + \big( x-y-2\big)^2 + \big(x+2y-3 \big)^2 = 3(x-2)^2+6(y-\frac{1}{2})^2 +\frac{1}{2}$$ the minimun is $\frac{1}{2}$

$\endgroup$
2
$\begingroup$

hint:

$a = y-x+1, b = x-y-2, c = x+2y-3$, then find an equation in $a,b,c$ and use CS inequality.

$\endgroup$
1
$\begingroup$

since $$f=(-x+y+1)^2+(x-y-2)^2+(x+2y-3)^2\ge (-x+y+1)^2+(x-y-2)^2$$ and Use Cauchy-Schwarz inequality we have $$[(-x+y+1)^2+(x-y-2)^2][1^2+1^2]\ge (-x+y+1+x-y-2)^2=1$$ so $$f\ge \dfrac{1}{2}$$

$\endgroup$
0
$\begingroup$

If you expand the expression you see there are no cross terms, so it boils down to separately minimizing a function of $x$ and a function of $y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.