Steady State
MAT161 / Lecture 12
Lecture 12 · Barnett ch. 4 · about 50 min

Optimization

The capstone of differential calculus for business: the closed-interval method for guaranteed absolute extrema, and constrained optimization by substitution, applied to profit maximisation and cost minimisation.

By the end you can
  • Distinguish local extrema from absolute (global) extrema
  • Apply the closed-interval method to find an absolute maximum or minimum
  • Set up and solve a constrained optimization problem using substitution
  • Confirm a constrained optimum with the second-derivative test

Local versus absolute extrema

The first- and second-derivative tests from the last two lessons find local extrema: turning points relative to their immediate neighbourhood. A firm usually cares about the absolute (global) best: the single largest profit or smallest cost over the entire feasible range of output.

Extreme Value TheoremTheoremFormula sheet →
f continuous on [a,b]  ⇒  f attains an absolute max and an absolute min on [a,b]f \text{ continuous on } [a, b] \;\Rightarrow\; f \text{ attains an absolute max and an absolute min on } [a, b]
[a,b][a, b]
a closed, bounded interval

Use it when you need to guarantee that an absolute maximum or minimum actually exists before searching for it, such as when output is capped by capacity.

The closed-interval method

Closed-interval methodProcedureFormula sheet →
1. Find all critical numbers of f in (a,b)2. Evaluate f at each critical number and at a,b3. The largest value is the absolute max; the smallest is the absolute min1.\ \text{Find all critical numbers of } f \text{ in } (a,b) \qquad 2.\ \text{Evaluate } f \text{ at each critical number and at } a, b \qquad 3.\ \text{The largest value is the absolute max; the smallest is the absolute min}

Use it when output or another decision variable is restricted to a closed interval, such as a factory's minimum and maximum capacity.

Worked example · Finding the absolute maximum profit0/6

Profit is P(x)=−x3+15x2−48x+200P(x) = -x^3 + 15x^2 - 48x + 200, and the factory can produce between 0 and 10 units (a capacity limit). Find the absolute maximum profit on [0,10][0, 10]. (Critical numbers 2 and 8 were found in Lesson 10.)
Your turn

Same profit function P(x)=−x3+15x2−48x+200P(x) = -x^3 + 15x^2 - 48x + 200, restricted to [0,5][0, 5]. Find the absolute maximum profit on this interval.

Try it: unconstrained optimization

Constrained optimization by substitution

Many real decisions involve two variables linked by a constraint: a fixed volume, a fixed budget, a fixed area. The technique is to solve the constraint for one variable, substitute it into the objective function, and optimize the resulting one-variable function with the tools already built.

Constrained optimization by substitutionProcedureFormula sheet →
1. Write the objective function of two variables2. Write the constraint and solve it for one variable3. Substitute to get a one-variable function4. Optimize as usual1.\ \text{Write the objective function of two variables} \qquad 2.\ \text{Write the constraint and solve it for one variable} \qquad 3.\ \text{Substitute to get a one-variable function} \qquad 4.\ \text{Optimize as usual}

Use it when a cost or area depends on two design variables linked by a fixed relationship, such as a fixed volume or a fixed budget.

Worked example · Minimizing the cost of an open storage bin0/6

A firm needs an open-top storage bin with a square base, holding exactly 9 cubic metres. The base costs 3 TL per square metre; the four sides cost 4.5 TL per square metre. Let x be the base's side length and h its height. Find x and h that minimise total material cost.
Your turn

Using the same setup, what is the minimum cost, in TL, if the required volume is instead 72 cubic metres, with the same 3 TL and 4.5 TL rates? (Use C(x)=3x2+18×72x=3x2+1296xC(x) = 3x^2 + \dfrac{18 \times 72}{x} = 3x^2 + \dfrac{1296}{x}.)
Check your understanding

In a constrained optimization problem, after substituting the constraint into the objective function, what is the next step?

Exam practice

Exam question 1

Revenue is R(x)=80x−0.4x2R(x) = 80x - 0.4x^2 for xx in [0,120][0, 120] (a capacity limit). Find the critical number of R.
Exam question 2

For the same R(x) = 80x − 0.4x², find the absolute maximum revenue on [0, 120] by comparing R(0), R(100), and R(120).
Exam question 3

A box with a square base and open top must hold 8 cubic metres. The base costs 2 TL/m², and each of the four sides costs 1 TL/m². Using C(x)=2x2+32xC(x) = 2x^2 + \dfrac{32}{x}, find the side length x that minimises cost.
Exam question 4

Why must you check the endpoints of a closed interval, even after finding a local maximum inside it?

Summary and review

Review deck · 7 cards0/7 mastered