Answer to Question #275097 in Calculus for Nemar

Question

It's time to tidy up your work desk. you are given 27cm^2 of cardboard to build a rectangular box without a lid to store small electronic components. By using the knowledge of partial derivative, determine the maximum volume of this box

Answer 1

x, y, z are sides of rectangular

surface area:

$A=xy+2xz+2yz=27$ cm²

height of box:

$z=\frac{27-xy}{2(x+y)}$

volume of box:

$V=xyz=xy\frac{27-xy}{2(x+y)}$

$V_x=\frac{(27y-2xy^2)(x+y)-27xy+x^2y^2}{2(x+y)^2}=\frac{27y^2-x^2y^2-2xy^3}{2(x+y)^2}=0$

$V_y=\frac{(27x-2yx^2)(x+y)-27xy+x^2y^2}{2(x+y)^2}=\frac{27x^2-x^2y^2-2yx^3}{2(x+y)^2}=0$

$27y^2-x^2y^2-2xy^3=0$

$27x^2-x^2y^2-2yx^3=0$

$27y^2-27x^2-2xy^3+2yx^3=0$

$27(y^2-x^2)+2xy(x^2-y^2)=0$

$27-2xy=0$

$x=27/(2y)$

$27y^2-729/4-27y^2=0$

$x^2-y^2=0$

$x=y$

$27x^2-x^4-2x^4=0$

$x^2(27-3x^2)=0$

$x=y=3$

$V_{max}=9\frac{27-9}{2(3+3)}=13.5$ cm³