NW Food Ltd needs to store 500 m3 (cubic meters) of wheat in identical rectangular boxes. The top and bottom of each box will be made from a type of material which costs £1.00/m2 (£1.00 per square meter). The four sides of each box will be made from another type of material which costs £1.50/ m2. It costs £0.20 to assemble each box. What should be the dimensions and the total number of boxes to minimise the cost of storing the wheat? You are required to construct a mathematical program to represent the problem. What type of mathematical program is this (linear/non-linear, constrained/non-constrained, single/multi variable, integer, etc.)?

I've attempted to solve it

Assuming, dimensions are: length of base =x, breadth of base =y, height of box =z

Volume of box = xyz

number of boxes required = 500/xyz

Area of Base+Top = 2xy

Area of Sides = 2xz + 2yz

Cost per box = (2xy + 3xz + 3yz +0.2)

Total Cost = (500/xyz)(2xy + 3xz + 3yz +0.2) = (1000/z) + (1500/y) + (1500/x) + (100/xyz)

I've taken the partial derivative of the total with respect to x, y and z and equate them to 0.

However, the results are odd. for the partial derivate with respect to x is (-1500/x^2) - (100/yzx^2) = 0, which can't really be solved.

Where have i gone wrong?


1 Answer 1


Your objective shows that the partial derivatives are all negative, i.e., the costs go down as $x$, $y$ or $z$ increases. This makes sense: if you have more than one box and you put them next to each other, you pay for the sides that are 'in between', on top of extra assemble costs. So, you should use a single large box. The problem then becomes to maximize $20xy + 3xz + 3yz + 0.2$ such that $xyz=500$. The method of Lagrange with the Lagrangian $L(x,y,z,\lambda) = 20xy + 3xz + 3yz + 0.2 - \lambda(xyz-500)$ gives: $$20y+3z-\lambda yz = 0$$ $$20x+3z-\lambda xz = 0$$ $$ 3x+3y-\lambda xy = 0$$ $$xyz = 500$$ The solution is: $$ \lambda = 2\left(\frac{3}{5}\right)^{2/3} $$ $$ x = y = 3^{1/3} 5^{2/3}$$ $$ z = 20 \left(\frac{5}{3}\right)^{2/3}$$


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