Answer to Question #123070 in Geometry for Jamshaid Hameed

Question

Joe has 650 cm^2 of tin. He wants me to make a tin of can that can hold as much juice as possible. What dimensions should he use? What is the largest volume this can could have?

Answer 1

$V^2=x^2\times y^2\times z^2=x^2\times y^2\times (d^2-x^2-y^2);\\d/dx: 2\times x \times y^2 \times (d^2 -\\- x^2 - y^2) - x^2\times y^2\times 2x = 0;\\ 2\times x\times y^2\times (d^2 - x^2 - y^2 - x^2) = 0;\\ 2\times x^2 + y^2 = d^2;\;1)\\ d/dy:2\times y\times x^2\times(d^2 - x^2 - y^2) -\\- x^2 \times y^2 \times 2y = 0;\\ 2\times y\times x^2\times (d^2 - x^2 - y^2 - y^2) = 0;\\ x^2 + 2y^2 = d^2\;2)\\x^2-y^2=0;x=y;\\take \;equation\; 1:\\2\times x^2+y^2=d^2\; and\;x=y=\frac{d}{\sqrt{3}};\\x^2+y^2+z^2=d^2;\;then\;x=y=z;\\The \;largest\; volume \;of\; a\; rectangular\\ parallelepiped\; is \;under\; the\; condition\; a=b=c;\\V=a^3;S=a^2+4\times a^2=650;\\5\times a^2=650;a^2=130;\\V=130\times\sqrt{130}sm^3;\\Answer:the\; largest \;volume\;\, is\, V=130\times\sqrt{130}sm^3$