Optimization..minimizing cost of container

Question

A rectangular storage container with an open top is said to have a volume of 10m^2. The length of the base is twice the width of the base . Material for the base costs $10 per square metre. Material for the sides costs $6 per square metre. Find the cost of materials for the cheapest such container.

Answer 1

Volume= width*length*height.  Let x = width, y = height.  Since V = 10 m³:

 

     10 = x*2x*y = 2x²y

     5/x² = y  (This is your substitution equation)

 

The cost of the container is:

 

     C = (area of sides)*(cost per unit area of sides) + (area of base)*(cost per unit area of base)

     C = (2*2xy + 2*xy)*($6)  +  (x*2x)*($10)

     C = 36xy  +  20x²

 

The above equation is the cost equation in 2 variables (x and y).  To convert it to one variable, substitute 5/x² for y from the substitution equation:

 

     C = 36xy + 20x²

     C = 36x(5/x²) + 20x²

     C = 180/x + 20x²

 

To find the dimensions of the box with the minimum cost:

1. Take the first derivative of C wrt x, dC/dx
2. Set it to zero
3. Solve for x

x is the width.  The length is 2x, the height is 5/x².