Solution:

First we do parts a,b,c by the usual method. We use the hand calculation technique, rather than Maple routines.

Now augment A by B

Apply Gauss elimination

Gausselim is done. We carry out Gaussjord, since it is so short!

Now read of the and . We use the variables p,c,w,h and remark that the subscript p in is from the standard "particular solution" notation and not the "peanut" variable. We get , with being free. So the null space is:

>

and the particular solution is:

Add amd to get the general solution

Now to the last parts.
Part d.

By looking at the equations in the general solution, we see that the particular solution [p,c,w,h] = [30/7,40/7,0,0] yields the maximum peanuts and cashews, since increasing w or h even a little will decrease p and c.

Part e.
Note that all components of have to be nonnegative. This gives a maximum of divided by or from the top entry, while divided by or from the second. Thus the maximum amount of walnuts is 8 pounds! Similar calculations yield the minimum of and 20 or 6 as the largest amount of hazelnuts.

Part e.

We sketch the regions showing the entries to be nonnegative. Each entry gives one side of a line, together the shaded region is it!

Note that these questions ask you to use the Linear Algebra with the help of common sense. It is an essential part of learning to be able to extend your knowledge in this way.