7.4 Graphical Solutions to Linear Optimization Problems

To consider how the objective function connects, suppose we considered all the possible production combinations that gave a profit of [latex]P = $3000[/latex], so that [latex]3000 = 30b + 40p[/latex]. That set of combinations would form a line in the graph. Doing the same for a profit of $5000 and $6500 would give additional lines. Graphing those on top of our feasible region, we see a pattern:

Notice that all the constant-profit lines are parallel, and that in general the profit increases as we move up to upper right. Notice also that for a profit of $5000 there are some production levels inside the feasible region for that profit level, but some are outside. That means we could feasibly make $5000 profit by producing, for example, 167 basic items and no premium items, but we can’t make $5000 by producing 125 premium items and no basic items because that falls outside our constraints.

The solution to our linear programming problem will be the largest possible profit that is still feasible. Graphically, that means the line furthest to the upper-right that still touches the feasible region on at least point. That solution is the one below:

This profit line touches the feasible region where [latex]b = 195[/latex] and [latex]p = 0[/latex], giving a profit of [latex]P[/latex] = 30(195) + 40(0) = $5850.

Notice that this is slightly larger than the profit that would be made by completely utilizing all staffing at [latex]b = 120[/latex], [latex]p = 50[/latex], where the profit would be $5600.

A health-food business would like to create a high-potassium blend of dried fruit in the form of a box of 10 fruit bars. It decides to use dried apricots, which have 407 mg of potassium per serving, and dried dates, which have 271 mg of potassium per serving. The company can purchase its fruit in bulk for a reasonable price. Dried apricots cost $9.99/lb. (about 3 servings) and dried dates cost $7.99/lb. (about 4 servings). The company would like the box of bars to have at least the recommended daily potassium intake of about 4700 mg, and contain at least 1 serving of each fruit. In order to minimize cost, how many servings of each dried fruit should go into the box of bars?

We begin by defining the variables. Let
[latex]x[/latex] = number of servings of dried apricots
[latex]y[/latex] = number of servings of dried dates

We next work on the objective function.

For apricots, there are 3 servings in one pound. This means that the cost per serving is $9.99/3 = $3.33. The cost for [latex]x[/latex] servings would thus be [latex]3.33x[/latex].

For dates, there are 4 servings per pound. This means that the cost per serving is $7.99/4 = $2.00. The cost for [latex]y[/latex] servings would thus be [latex]2.00y[/latex].

The total cost, [latex]C[/latex], for apricots and dates would be [latex]C = 3.33x + 2.00y[/latex]

Normally we would have constraints [latex]x\geq0[/latex] and [latex]y\geq0[/latex] since negative servings can’t be used. But in this case, we’re further restricted. In words:

• There must be at least 1 serving of each fruit
• The product must contain at least 4700 mg of potassium

Mathematically,

• Since there must be at least 1 serving of each fruit, [latex]x\geq1[/latex] and [latex]y\geq1[/latex]
• There are [latex]407x[/latex] mg of potassium in [latex]x[/latex] servings of apricots and [latex]271y[/latex] mg of potassium in [latex]y[/latex] servings of dates. The sum should be greater than or equal to 4700 mg of potassium, or [latex]407x + 271y\geq4700[/latex]

Thus we have,

Objective function:

[latex]C = 3.33x + 2.00y[/latex]

Constraints:

[latex]\begin{align*} 407x + 271y&\geq4700\\ x&\geq1 , y\geq1 \end{align*}[/latex]

We graph the constraints and shade the feasible region:

The region is unbounded, but we will be able to find a minimum still. We can see there are two corner points.

The one in the upper left is the intersection of the lines [latex]407x + 271y = 4700[/latex] and [latex]x = 1[/latex]. Solving for the intersection using substitution:

\begin{align*}
407(1) + 271y &= 4700\\
y &\approx 15.8\\
\end{align*}

Point: [latex](1, 15.8)[/latex]

The one in the lower right is the intersection of the lines [latex]407x + 271y = 4700[/latex] and [latex]y = 1[/latex].

\begin{align*}
407x + 271(1) &= 4700\\
x &\approx 10.9\\
\end{align*}

Point: [latex](10.9, 1)[/latex]

Testing the objective function at each of these corner points:

The company can minimize cost by using 1 serving of apricots and 15.8 servings of dates.

A factory manufactures chairs and tables, each requiring the use of three operations: Cutting, Assembly, and Finishing. The first operation can be used at most 40 hours; the second at most 42 hours; and the third at most 25 hours. A chair requires 1 hour of cutting, 2 hours of assembly, and 1 hour of finishing; a table needs 2 hours of cutting, 1 hour of assembly, and 1 hour of finishing. If the profit is $20 per unit for a chair and $30 for a table, how many units of each should be manufactured to maximize profit?

We begin by defining the variables. Let
[latex]c[/latex] = number of chairs made
[latex]t[/latex] = number of tables made

The profit, [latex]P[/latex], will be [latex]P = 20c + 30t[/latex].

For cutting, [latex]c[/latex] chairs will require [latex]1c[/latex] hours and [latex]t[/latex] tables will require [latex]2t[/latex] hours. We can use at most 40 hours, so [latex]c + 2t \leq40[/latex].

For assembly, [latex]c[/latex] chairs will require [latex]2c[/latex] hours and [latex]t[/latex] tables will require [latex]1t[/latex] hours. We can use at most 42 hours, so [latex]2c + t\leq42[/latex].

For finishing, [latex]c[/latex] chairs will require [latex]1c[/latex] hours and [latex]t[/latex] tables will require [latex]1t[/latex] hours. We can use at most 25 hours, so [latex]c + t\leq25[/latex].

Since we can’t produce negative items, [latex]c\geq 0, t \geq 0[/latex].

Graphing the constraints, we can see the feasible region.

There are five corner points for this region.

Point 1: In the lower left, where [latex]t = 0[/latex] crosses [latex]c = 0[/latex].

Point: [latex](0, 0)[/latex]

Point 2: In the upper left, where [latex]c = 0[/latex] crosses [latex]c + 2t = 40[/latex].

Using substitution, [latex]0 + 2t = 40[/latex], so [latex]t = 20[/latex].

Point: [latex](0, 20)[/latex]

Point 3: In the lower right, where [latex]t = 0[/latex] crosses [latex]2c + t = 42[/latex].

Using substitution, [latex]2c + 0 = 42[/latex], so [latex]c = 21[/latex].

Point: [latex](21, 0)[/latex]

Point 4: Where [latex]c + 2t = 40[/latex] crosses [latex]2c + t = 42[/latex].

We can solve this as a system using any techniques we know. We could solve the second equation for [latex]c[/latex], giving [latex]c = 25 - t[/latex], then substitute into the first equation:

[latex]\begin{align*} (25 - t) + 2t &= 40\\ 25 + t &= 40\\ t &= 15\\ \end{align*}[/latex]

Then [latex]c = 25 – 15 = 10[/latex].
Point: [latex](10, 15)[/latex]
Point 5: Where [latex]2c + t = 42[/latex] crosses [latex]c + t = 25[/latex].

We can solve this as a system using any techniques we know. Using a different technique this time, we could multiply the bottom equation by -1 then add it to the first:

[latex]\begin{align*} 2c + t &= 42\\ -c - t &= -25 & {\textrm{equation 1- equation 2}}\\ c &= 17\\ \end{align*}[/latex]

Then using [latex]c + t = 25[/latex], we have [latex]17 + t = 25[/latex], so [latex]t = 8[/latex].
Point: [latex](17, 8)[/latex]

Testing the objective function at each of these corner points:

The profit will be maximized by producing 10 chairs and 15 tables.