Find the maximum value of $ax+ by$

Question

$$ x^2 + xy + y^2 = t^2 $$

Find the maximum value of $ax + by$

One way of doing this is substituting

$ x = r \cos w $ and $ y= r \sin w $

Then using calculus we can find the maximum value but this is a very lengthy process

So I wanted to know if there is a shorter way of doing this

Answer 1

you can use the Lagrange Multiplier Method $$f(x,y,\lambda)=ax+by+\lambda(x^2+xy+y^2-t^2)$$

Answer 2

It might be pretty lengthy too, but another way is using a Langrange multiplier.

You would get need to solve $(a,b)=\lambda (2x+y,x+2y)$ with the constraint $x^2 + xy + y^2 = t^2$, so after solving the linear equations you would get $x=\frac{2a-b}{3\lambda}$ and $y=\frac{-a+2b}{3\lambda}$, then plugging this in the contraint would give the needed $\lambda$.

So if you do it by hand, it is also lengthy (unless of course if you know the inverse of a $2\times 2$-matrix by heart, which isn't difficult to find), but using some program to solve linear equations you are done in no time.

Answer 3

$$(ax+by)^2\leq\frac{4}{3}(a^2-ab+b^2)(x^2+xy+y^2)$$ it's $$((a-2b)x+(2a-b)y)^2\geq0.$$ The equality occurs for $(a-2b)x+(2a-b)^2y=0$ and $x^2+xy+y^2=t^2.$

Thus, $$\max\limits_{x^2+xy+y^2=t^2}(ax+by)=\frac{2}{\sqrt3}|t|\sqrt{a^2-ab+b^2}.$$