Two-Product Profit With Shared Capacity

Modeling Profit With a Shared Capacity

When two products use the same limited production resource, choosing how much of each to make is a constrained optimization problem. Producing more of one product may leave less capacity for the other, so maximizing each product's profit separately will not generally maximize total profit.

Let $x$ and $y$ denote the quantities of two products. Suppose their revenue functions are $R_1(x)$ and $R_2(y)$, and their cost functions are $C_1(x)$ and $C_2(y)$. If producing one unit of the first product uses $a$ units of capacity and one unit of the second uses $b$, the shared capacity limit $K$ gives the constraints $x\ge 0$, $y\ge 0$, and $ax+by\le K$, where $a,b,K>0$.

Definition: The profit function is revenue minus cost. In this separable two-product model, $$ \pi(x,y)=R_1(x)+R_2(y)-C_1(x)-C_2(y). $$ The feasible production region is $$ \{(x,y):x\ge 0,\ y\ge 0,\ ax+by\le K\}. $$

The feasible region is a closed triangle, so it is compact. If profit is continuous, it attains a global maximum somewhere in that region. The maximum may lie inside the triangle, on the shared-capacity edge, on an axis, or at a corner. A reliable solution method must account for all these possibilities.

Which Candidates Must Be Checked?

At a point with $x>0$, $y>0$, and $ax+by<K$, small changes in either production quantity remain feasible. If $\pi$ is differentiable, an interior local maximum must therefore satisfy $\pi_x=0$ and $\pi_y=0$.

If instead $x>0$, $y>0$, and $ax+by=K$, the capacity constraint is active. Along this edge, a change in $x$ requires a corresponding change in $y$. A differentiable maximum in the interior of this edge must satisfy the Lagrange multiplier condition below. This condition applies to positive interior points of the capacity edge; it cannot be imposed without modification at a corner where an axis constraint is active too.

Necessary-condition checklist: For a differentiable profit function on the triangular feasible region, check feasible interior stationary points; stationary points in the positive interior of the capacity edge; and maxima on each axis segment, including its endpoints. The endpoints include the corners, which should be evaluated directly.

On the capacity edge, $y=(K-ax)/b$. The derivative of profit along the edge is therefore $\pi_x-(a/b)\pi_y$. At an interior maximum on that edge, this derivative is zero, giving the equivalent conditions

\pi_x-\frac{a}{b}\pi_y=0 \qquad\Longleftrightarrow\qquad \nabla\pi=\lambda(a,b)

for some scalar $\lambda$. The equivalence follows because the edge condition is $b\pi_x=a\pi_y$: the gradient is parallel to the capacity normal $(a,b)$. The endpoint cases are excluded from this argument because feasible changes there are restricted by both capacity and nonnegativity.

Worked Example: A Capacity-Edge Corner Is Best

Worked Example: A Three-Hour Production Limit

Product A uses two hours per unit and product B uses one hour per unit. There are three production hours available, so $2x+y\le 3$. Suppose revenues and costs, in dollars, give the profit function $\pi(x,y)=16x-2x^2+18y-y^2$. Find the maximum profit.

First, consider a possible maximum with positive production of both products and unused capacity. Its derivatives must both vanish:

\pi_x=16-4x=0 \Longrightarrow x=4, \qquad \pi_y=18-2y=0 \Longrightarrow y=9.

This point uses $2(4)+9=17$ hours, so it is infeasible. There is no feasible slack interior stationary point.

On the capacity edge, write $y=3-2x$, with $0\le x\le 3/2$. Differentiating the profit along that edge gives

\frac{d}{dx}\pi(x,3-2x) =\pi_x-2\pi_y =(16-4x)-2\bigl(18-2(3-2x)\bigr) =-8-12x.

This derivative is negative throughout the interval, so profit decreases as $x$ increases along the capacity edge. Its maximum on that edge is at $x=0,\ y=3$, where $\pi(0,3)=54-9=45$ dollars.

Check the other axis segment. When $y=0$, capacity requires $0\le x\le 3/2$, and the profit derivative is $16-4x\ge 10>0$ throughout that interval. Its maximum is at $x=3/2$, giving $\pi(3/2,0)=24-9/2=39/2=19.5$ dollars. When $x=0$, the edge $0\le y\le 3$ has derivative $18-2y\ge 12>0$, so its maximum is again at $y=3$.

Thus the best feasible point is $(0,3)$, with profit $45$ dollars. At that point, $\nabla\pi=(16,12)$, while the capacity normal is $(2,1)$; these are not parallel because $16/2=8\ne 12/1=12$. There is no contradiction: this maximizer is also on the nonnegativity boundary $x=0$, so the capacity-edge multiplier condition for positive $x,y$ does not apply there. Directly checking the corner and adjacent feasible edges gives the correct conclusion.

Worked Example: An Interior Point of the Capacity Edge

Worked Example: Comparing an Edge Candidate With the Boundaries

Suppose $\pi(x,y)=30x-x^2+24y-2y^2$, with $x,y\ge 0$ and $x+2y\le 10$. Find the global maximum.

On the positive interior of the capacity edge, the multiplier equations are $30-2x=\lambda$ and $24-4y=2\lambda$. Combining them gives $24-4y=60-4x$, hence $x-y=9$. Together with $x+2y=10$, this yields $y=1/3$ and $x=28/3$. Both quantities are positive, and the capacity is exactly used. The multiplier is $\lambda=30-2(28/3)=34/3$; the second equation checks since $24-4/3=68/3=2\lambda$.

The profit at this point is

\pi\left(\frac{28}{3},\frac13\right) =30\left(\frac{28}{3}\right)-\left(\frac{28}{3}\right)^2 +24\left(\frac13\right)-2\left(\frac13\right)^2 =\frac{2520-784+72-2}{9} =\frac{602}{3}.

For comparison, on the axis $y=0$, $0\le x\le10$, and the profit $30x-x^2$ is increasing because its derivative $30-2x$ is at least $10$. Its largest value there is $200$, at $(10,0)$. On the axis $x=0$, $0\le y\le5$, and $24y-2y^2$ is increasing because its derivative $24-4y$ is at least $4$. Its largest value there is $70$, at $(0,5)$.

To verify global optimality, let $z=(x,y)$ be any feasible point. Expanding the quadratic around $p=(28/3,1/3)$ gives

\pi(x,y) =\pi(p)+\frac{34}{3}\left(x-\frac{28}{3}\right) +\frac{68}{3}\left(y-\frac13\right) -\left(x-\frac{28}{3}\right)^2 -2\left(y-\frac13\right)^2.

The two linear terms combine to $(34/3)[(x+2y)-10]$, which is nonpositive for every feasible point. The squared terms are also nonpositive after their minus signs. Consequently $\pi(x,y)\le\pi(p)=602/3$ throughout the feasible region. The candidate is the global maximum, with profit $602/3$ dollars. This argument verifies the boundary cases as well as interior points.

Worked Example: Capacity Need Not Be Fully Used

Worked Example: A Feasible Slack Optimum

Let $\pi(x,y)=16x-x^2+12y-2y^2$, with $x,y\ge0$ and $2x+y\le20$. The stationary equations give $16-2x=0$, so $x=8$, and $12-4y=0$, so $y=3$. The capacity used is $2(8)+3=19$, which is less than $20$, so this is a feasible slack point.

Its profit is $16(8)-8^2+12(3)-2(3^2)=128-64+36-18=82$. To verify it is globally optimal, complete the squares:

\pi(x,y)=64-(x-8)^2+18-2(y-3)^2 =82-(x-8)^2-2(y-3)^2\le82.

Equality holds only at $(8,3)$. Substitution confirms that this point is feasible and attains the bound, so it is the global maximum. The shared capacity is not automatically used in full: producing more would reduce profit in this model.

Why the Boundary Checks Matter

The first example illustrates a common error: applying the capacity multiplier equation to every point where capacity is used up. At a corner such as $(0,3)$, feasible movements must respect both $2x+y\le3$ and $x\ge0$. The usual one-dimensional capacity-edge derivative condition assumes that small changes in either direction along that edge are available, which is why it is necessary only at a positive interior point of the edge. At an endpoint, evaluate the corner and check the adjacent boundary segments instead.

A second error is to solve the multiplier equations and stop. Those equations identify candidates, not automatically global maxima. A candidate may be infeasible, may be a minimum along the edge, or may lose to an axis point. A complete solution either compares all relevant candidates and boundary maxima or supplies a valid global argument, such as a concavity bound. In each example, the feasibility check is part of the calculation, not an optional final detail.

Write profit and constraints.
Combine the two revenues and costs, and state nonnegativity and shared capacity explicitly.

Find feasible interior candidates.
For a point with slack capacity and positive production, solve $\nabla\pi=\mathbf 0$.

Check the positive capacity edge.
Use the multiplier equations or differentiate profit along the edge; also evaluate its endpoints.

Check both axis segments.
Optimize profit on each segment, including its endpoints, then compare feasible candidates or prove a global bound.

Check Your Understanding

Use the definitions and boundary checks in this tutorial to answer the following questions.

For $ax+by\le K$, how can you parameterize the capacity edge, and what interval of $x$-values is feasible on it?
Why is the multiplier condition for the capacity edge restricted to points with $x>0$ and $y>0$?
What candidates must be checked when maximizing a differentiable profit function over the triangular feasible region?
For $\pi(x,y)=10x-x^2+8y-y^2$ with $x,y\ge0$ and $x+y\le5$, find the positive interior capacity-edge stationary point, if one exists.
Give one way to verify that a feasible candidate is a global maximum rather than merely a stationary point.