Calculus on equations with 2 variables

Question

Suppose $x$ and $y$ are real numbers such that $$x^2 + y^2 - xy = 10.$$ What is the maximum value of $60(x^2 + y^2 + xy)$?

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I transposed the variables and used substitution and obtained $1200$ as the answer. I just want to be definite if my answer is correct.

Answer 1

Using Lagrange multipliers method: $$L=60(x^2+y^2+xy)+t(10-x^2-y^2+xy).$$ $$\begin{cases} L_x=120x+60y-2xt+ty=0 \\ L_y=120y+60x-2yt+tx=0 \\ L_t=10-x^2-y^2+xy=0\end{cases} \Rightarrow\begin{cases} x_{1,2}=y_{1,2}=\pm \sqrt{10} \\ x_{3,4}=-y_{3,4}=\pm \sqrt{\frac{10}{3}}\end{cases}$$ Hence: $$f(x_1,y_1)=f(x_2,y_2)=1800 - max;$$ $$f(x_3,y_3)=f(x_4,y_4)=200 - min.$$

Answer 2

I got $1800$ because $$60(x^2+xy+y^2)\leq1800\cdot\frac{x^2-xy+y^2}{10}$$ it's $$3(x^2-xy+y^2)\geq x^2+xy+y^2,$$ which is $$(x-y)^2\geq0.$$ The equality occurs for $x=y$, which says that $1800$ is a maximal value.