","comment":{"@type":"Comment","text":" Find the perimeter in terms of $x$ and $y$ and equal it to $750$. It will give us the relation between $x$ and $y$. Now find the area in terms of $x$ and $y$. Find the $x$ where $\\dfrac{dA}{dx}=0$ which gives the value of ${{A}_{\\max }}$."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $${{A}_{\\max }}=\\dfrac{{{\\left( 375+2\\pi \\right)}^{2}}}{\\pi }$$ ","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $${{A}_{\\max }}=\\dfrac{{{\\left( 375+2\\pi \\right)}^{2}}}{2\\pi }$$ ","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $${{A}_{\\max }}=\\dfrac{{{\\left( 375+2\\pi \\right)}^{2}}}{\\pi }$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $${{A}_{\\max }}=\\dfrac{{{\\left( 375+\\pi \\right)}^{2}}}{2\\pi }$$","position":2,"answerExplanation":{"@type":"Comment","text":" The length of the material is equal to $$750 ft$$. So, the perimeter of the cricket ground is equal to $$750 ft$$. $$ \\\\ \\Rightarrow 750=\\pi x+\\pi x+2(x+y) \\\\ \\Rightarrow y=375-\\pi (x-1)....(1) \\\\ $$ Let us assume the area is equal to $$A$$. $$ \\\\ \\Rightarrow A=\\dfrac{\\pi {{x}^{2}}}{4}+\\dfrac{\\pi {{x}^{2}}}{4}+xy....(2) \\\\ $$ Now substitute equation $$\\left( \\text{2} \\right)$$ in equation $$\\left( \\text{1} \\right)$$, then we get $$ \\\\ \\Rightarrow A=\\dfrac{\\pi {{x}^{2}}}{4}+\\dfrac{\\pi {{x}^{2}}}{4}+x\\left( 375-\\pi \\left( x-1 \\right) \\right) \\\\ \\Rightarrow A=(375+\\pi )x+\\dfrac{-\\pi {{x}^{2}}}{2} \\\\ $$ The value of $$x$$ where $$\\dfrac{dA}{dx}=0$$ gives the value of $${{A}_{\\max }}$$. $$ \\\\ \\Rightarrow \\dfrac{dA}{dx}=0 \\\\ \\Rightarrow \\dfrac{d}{dx}\\left( (375+\\pi )x+\\dfrac{-\\pi {{x}^{2}}}{2} \\right)=0 \\\\ \\Rightarrow x=\\dfrac{375+\\pi }{\\pi }...(3) \\\\ $$ Now substitute equation $$\\left( \\text{3} \\right)$$ in equation $$\\left( \\text{2} \\right)$$ $$ \\\\ \\Rightarrow {{A}_{\\max }}=(375+\\pi )\\left( \\dfrac{375+\\pi }{\\pi } \\right)+\\dfrac{-\\pi {{\\left( \\dfrac{375+\\pi }{\\pi } \\right)}^{2}}}{2} \\\\ \\Rightarrow {{A}_{\\max }}=\\dfrac{{{\\left( 375+\\pi \\right)}^{2}}}{2\\pi } \\\\ $$ So, it is clear that $${{A}_{\\max }}=\\dfrac{{{\\left( 375+\\pi \\right)}^{2}}}{2\\pi }$$.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}