2) A closed rectangular box with square base and a volume of 21 cubic feet is to

Question

2) A closed rectangular box with square base and a volume of 21 cubic feet is to be constructed using two different types of materials. The top is made of metal costing $5 per square foot and the remainder of plastic costing $2 per square foot. Let x be the length of the side of the base and let y be the height of the box. Consider the problem of finding the dimensions that would yield the smallest possible cost. (a) Determine the objective function. (b) Determine the constraint equation. (c) Express the quantity to be minimized in terms of x. (d) Find the dimensions of the box that minimize cost

Explanation / Answer

The cube has a squar shaped base;
So dimensions of the cube : Length = x, breadth = x and height = y;
where x and y are in feet;
Material to be used = base + top + 4 walls = x*x + x*x + 4y*x

Cost of the top (metal) = 5 $ / sq ft and cost of remaining material = 2$ / sq ft;

For smalles possible cost:
Objective function: 5 (area of top) + 2 (area of remaining surfaces)
= 5 (x2) + 2 (x2 + 4xy)
= 7x2 + 8xy

Objective function is : Minimise 7x2 + 8xy;

Constraint: Volume constraint = x*x*y = x2y = 21;

So y = 21/x2
Substituting this in equation of objective function we get
Minimise 7x2 + 8x(21/x2) = 7x2 + 168/x;

To minimise cost, we are to take derivative of the objective function and equate to zero to find the value of 'x';
derivative of objective function = 14x -168/x2 =0
14x = 168/x2
x3 = 12
x= 121/3;

If x= 121/3 then y = 21/x2 = 21/ (121/3) 2 = 21/ 122/3;

Thus, for minimum cost, length = breadth = 121/3 and height = 21/122/3

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