Minimum and Maximum #1

Let $x+y=t$ and $x-y=s$ .

You see that $t+s>0$ and $t-s\geq0$ . Therefore, $t>0$ .

From $x^2-y^2=ts\geq1$ , we get $s\geq\dfrac{1}{t}$ .

$ax+ay=at$ , $bx-by=bs$ . Then $(a+b)x+(a-b)y=at+bs$ .

We want the left side to become $5x-3y$ .

$a=1,~b=4$ .

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Therefore,

$5x-3y=t+4s\geq t+\dfrac{4}{t}\geq2\cdot \sqrt{t\cdot\dfrac{4}{t}}=\boxed{4}$ ,

with equality achieved when $t=2,~s=\dfrac{1}{2}$ , which means $x=\dfrac{5}{4},~y=\dfrac{3}{4}$ .