Mohon bantuanyaaaaaaaa

[tex]\displaystyle K=2x+2y+\frac12\pi(2x)\\p=2x+2y+x\pi\\p=(2+\pi)x+2y\\\frac{p-(2+\pi)x}{2}=y\\\\L=2xy+\frac12\pi r^2\\L=2x\cdot\frac{p-(\pi+2)x}{2}+\frac12\pi x^2\\L=\frac{2px-2(\pi+2)x^2+\pi x^2}{2}\\L=\frac{2px-2\pi x^2-4x^2+\pi x^2}{2}\\L=\frac{2px-\pi x^2-4x^2}{2}\\\text{agar maksimum, maka }L'=0\\L'=\frac{2p-2\pi x-8x}{2}\\0=p-\pi x-4x\\(\pi+4)x=p\\\boxed{\boxed{x=\frac{p}{\pi+4}}}[/tex]