Optimization - find dimensions of the box with the largest volume?

Question

John has been asked to construct a decorative wooden box in the shape of a square-based rectangular prism. The top of the box, which will be constructed with exotic hand carved wood will cost \$75/m^s. The sides of the box will be made with same wood, minus the carving, so it will cost \$50/m^2. The bottom of the box, which will not be seen, will cost \$20/m^2. If John has exactly \$400 to make this box, what are the dimensions of the box with the largest volume that he can produce?

I do not understand this question, what is the formula and how do I do this? Anybody please help me this question. Thanks.

Answer 1

You want to maximize the volume $$V=xyz$$ subject to the constraint $$400=75xy+20xy+50(2)(xz+yz)=95xy+100(xz+yz)$$ or $$80=19xy+20xz+20yz$$ Using Lagrange Multipliers, we get $$(yz,xz,xy)=\lambda (19y+20z,19x+20z,20x+20y)$$ Now solve this for x,y,z to get the dimensions of the box and use that to find the volume.

Answer 2

Let $x$ be the size of the base and $y$ be the height of the box. The volume is $v=x^2y$. The cost is $ C=75x^2+200xy+20x^2=95x^2+200xy.$ Therefore we have to maximize $x^2y,$ subject to the constraint $95x^2+200xy=400.$ We solve for y as a function of $x$ to get $$ y=(400-95x^2)/200x$$ Upon simplifying and substituting in the formula for $v$ we find $$v=x^2(400-95x^2)/200x =2x-(19/40)x^3.$$ $v'=0$ implies $x=\sqrt {80/57}.$ substituting in the expression for $y$ we get $y=19\sqrt {57/5}.$ therefore the maximum volume is $$v_{max} =1444\sqrt {57/5}$$