let y equal the number of type 2 fuse packages.
your objective function is to minimize the cost.
that equation is cost = 1 * x + .8 * y
your constraint functions are based on the following table:
type package #5A fuses # 10A fuses # 15A fuses
type 1 8 4 4
type 2 2 4 6
total required >= 16 >=20 >= 24
since x represents the number of type 1 packages and y represents the number of type 2 packages, your constraint equations become:
8x + 2y >= 16
4x + 4y >= 20
4x + 6y >= 24
you will want to graph these equations and then find the region of feasibility and then look at the corners of the graph which is the intersection points of each of the lines and then determine the minimum cost solution.
first to graph.
solve for y in each equation.
first equation gets you y >= (16-8x) / 2
second equation gets you y >= (20 - 4x) / 4
third equation gets you y = (24 - 4x) / 6
you will graph the equality portion of these equation to get the lines.
you will then use the inequality portion of these equations to get the area of feasibility.
let's take a look.
your graph will look like this:
your region of feasibility will be on or above all three of those lines.
your minimum / maximum cost points will be at the intersection of the lines of the region of feasibility.
those intersections are shown on the graph.
you need to evaluate your cost function at each of those points and then choose the point that has the minimum cost.
the points of intersection at the corners of the feasible region are:
(0,8) cost is equal to 0 * 1 + 8 * .8 which is equal to 6.4 dollars.
(1,4) cost is equal to 1 * 1 + 4 * .8 which is equal to 4.2 dollars.
(3,2) cost is equal to 3 * 1 + 2 * .8 which is equal to 4.6 dollars.
(6,0) cost is equal to 6 * 1 + 0 * .8 which is equal to 6.0 dollars.
looks like the minimum cost solution is 4.2 dollars.
this is when he buys 1 type 1 package and 4 type 2 packages.
let's see if that meets the requirements.
the purchase obtains the following:
number of type packages # 5A fuses # 10A fuses # 15A fuses 1 * type 1 8 4 4 4 * type 2 8 16 24 total 16 20 28
it meets the number of fuse requirements and has the lowest cost so this is the minimum cost solution.
note that only the intersections points that bound the region of feasibility are used. any other intersection points that do not bound the region of feasibility are not considered.