Question 783865
One bottom, and four sides.


Let x = length of side of bottom, a square shape.

Let 3 = length of side for dimension of the three-D figure, perpendicular to the square bottom, which is the height of the bin.  3 is how tall the bin.



Area of the bottom of the bin is {{{x^2}}}, and the cost is {{{5.40*x^2}}}.

Area for all four sides of the bin is {{{4*(3x)}}}, and the cost is {{{2.40*(4)(3x)}}}.


Wanted is cost be $63.
{{{5.40x^2+(2.40)4(3x)=63}}}
{{{5.4x^2+2.4*12x=63}}}
{{{5.4x^2+28.8x=63}}}
Multiply by 10 and divide by 6,
{{{9x^2+48x=70}}}, which is maybe the easiest equivalent of the equation to work with.
Put in more general form:
{{{highlight(9x^2+48x-70=0)}}}


Directly use general solution to quadratic equation.
Discrim, {{{(48)^2-4*9*(-70)=48^2+2520=4824}}}, which is {{{4*1201}}}
{{{x=(-48+sqrt(4*1201))/(2*9)}}}
{{{x=(-48+2*sqrt(1201))/(2*9)}}}
{{{highlight(x=(-24+sqrt(1201))/9)}}}