Giải thích các bước giải:

Ta có:

$2y+4x=60$

$\to y+2x=30$

$\to y=30-2x$

Diện tích khung là:

$S=\dfrac{x^2\sqrt3}4+xy$

$\to S=\dfrac{x^2\sqrt3}4+x(30-2x)$

$\to S=30x+\dfrac{x^2\left(-8+\sqrt{3}\right)}{4}$

$\to S=\dfrac{-8+\sqrt3}4\cdot (x+\dfrac4{-8+\sqrt3}\cdot 15)^2+\dfrac{900\left(8+\sqrt{3}\right)}{61}$

$\to S\le \dfrac{900\left(8+\sqrt{3}\right)}{61}$

Dấu = xảy ra khi $x+\dfrac4{-8+\sqrt3}\cdot 15=0$

$\to x=\dfrac4{-8+\sqrt3}\cdot 15\to y=30-2\cdot \dfrac4{-8+\sqrt3}\cdot 15$