optimised combined area of a rectangle and a square

Question

Consider a rectangle of dimensions $2x$ by $x$ and a square of dimensions $y$ by $y$. If the sum of the perimeters of the rectangle and square is $l$, find the value of $x$ and $y$ (in terms of $l$) that minimise the sum of areas of the rectangle and the square.

What I found on the internet are solutions where the rectangle and the square share one dimension.

Answer 1

$2x^2 + y^2 = A$

$6 x + 4 y = l$

so

$y = \frac{l - 6 x}{4}$

Thus $A = 2 x^2 + \left( \frac{l - 6 x}{4}\right)^2$

Expand... take the derivative with respect to $x$, set it to zero to find the minimizing (or maximizing) $x$.

Answer 2

You have given $$6x+4y=l$$ and to optimize the function $$f(x,y)=2x^2+y^2$$ with $$y=\frac{l-6x}{4}$$ you will get $$f\left(x,\frac{l-6x}{4}\right)=2x^2+\left(\frac{l-6x}{4}\right)^2$$ Can you proceed?