With exactly 2700 cm2 of cardboard we wish to construct a bo

With exactly 2700 cm2 of cardboard, we wish to construct a box (width, depth, height) that can contain a volume. We require the width to be double its depth. We
would like to maximize the volume the box can hold. Which values of x, y, and z fulfill our
objective. Show all calculations and screen shots from the input in Solver.

Solution

First, we must identify the variables, define the objectives and the constraints:
Three variables are described in this problem :
1. : box width (x> 0)
2. : box depth (y> 0)
3. : box height (z> 0)
The objective consists of maximizing the box volume. The objective function is
described by the expression V(x,y,z)=xyz
Two constraints are imposed:
The surface of available material is 2700 m2: 2xy+2yz+2xz= 2700
The requirements of the dimensions: x=2y

In a new Excel sheet, we will insert all this information using following the instructions:
1. To each variable we have to attribute a position on the worksheet : the cells B1,
B2 and B3 have been chosen to represent the variables x,yand z respectively;
2. Define the objective function : in B5, the objective is defined in function of the
variables B1,B2, and B3 ;
3. Define all constraints: The constraints are defined a bit differently than the
objective function. A constraint is a relation linking two expressions. For
example, 2xy+2yz+2xz= 2700
This requires equality between the expression 2xy+2yz+2xz and the expression
2700. In B7 and D7, each side of this relation is represented. In C7, we even identified
the nature of the relation linking B7 and D7. The second constraint x=2y is
represented by the cells B8, C8 and D8 in a similar manner.

The variables, the objective and the constraints having been inserted, we are ready to
solve the problem with the help of the Solver (Tools menu) :
1. Set objective: B5 contains the objective ;
2. To: we are looking to maximize the objective ;
3. By changing variable cells: B1, B2 and B3 represent the variables;
4. Subject to the constraints: by selecting the Add button, the two constraints of
the problem can be dictated to the Solver.
5. Use the checkbox: make unconstrained variables nonnegative (remember that
the variables representing the measurements of the sides of the box cannot be
negative).
Then, by clicking on Solve, the Solver will give you the solution to the problem.

Error messages
It is possible that the Solver sends the following message:

This can occur if the variables are initially on a fixed point of the objective function that
does not satisfy the requested criteria (maximum). In our example, if the cells B1, B2
and B3 are empty before questioning the Solver, they take on the value 0 by default.
Thus B1=0, B2=0, B3=0 is a fixed point of V=xyz,which explains the message error
you are receiving. To fix the situation, you need to modify the initial values of the
variables (by staying in the domain of possible solutions). For example, by giving the
cells B1, B2 and B3 the values 1, 1, and 1, the Solver will return the following solution:

In addition to the solution x= 30, y= 15, z= 20), we can observe that the
constraints are all respected…