# Unconstrained Multivariable Extrema

Business I · Mathematics II · https://tryals.app/learn/business-i/unconstrained-multivariable-extrema

## Finding Peaks and Pits

To find unconstrained extrema of $f(x,y)$, begin with **First-Order Conditions**: set $f_x = 0$ and $f_y = 0$. Their simultaneous solutions are **critical points**—our sole candidates for local extrema.

Next, use **Second-Order Conditions** via the **Hessian matrix** to classify each point:

$$H = \begin{pmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{pmatrix}$$

Compute the **determinant** $D = f_{xx} f_{yy} - (f_{xy})^2$ to reveal the local curvature.

> **Common pitfall:** Never check $f_{xx} < 0$ and $f_{yy} < 0$ separately. A large cross-term $f_{xy}$ can still yield a saddle point. Only the determinant test captures full surface interaction.

## The Hessian Verdicts

The determinant $D$ and $f_{xx}$ classify every critical point:

| $D$ Value | $f_{xx}$ Value | Verdict | Geometric Shape |
| --- | --- | --- | --- |
| $D > 0$ | $< 0$ | **Local maximum** | Curves down everywhere |
| $D > 0$ | $> 0$ | **Local minimum** | Curves up everywhere |
| $D < 0$ | Any | **Saddle point** | Mountain pass |
| $D = 0$ | Any | **Inconclusive** | Needs higher analysis |

**Economic application**: A firm choosing output $q$ and advertising $a$ to maximise profit $\pi(q,a)$ relies on FOCs for candidates and the Hessian to confirm a **local maximum**. If the Hessian yields a saddle point, the firm is at a pass where altering mix improves profit.

## Practice questions

7 of this lesson's 13 practice questions, with answers. The full set is in the app.

### 1. A saddle point is a maximum in one direction and a minimum in another. Why does a manager who tests only one direction confidently reach the wrong conclusion?

A. Because the first-order conditions fail at a saddle point
B. Because a one-dimensional test cannot see perpendicular fall
C. Because saddle points occur only in three or more variables
D. Because saddle points are unstable and move when tested

**Answer:** B. Because a one-dimensional test cannot see perpendicular fall

**Why:** Walk the pass one way and you are at a summit; turn ninety degrees and you are at the bottom of a valley, one cross-section cannot distinguish those cases. The Hessian determinant is the test that examines curvature in every direction simultaneously, which is why the SOC is a matrix condition rather than a sign.

Page: https://tryals.app/practice/business-i/unconstrained-multivariable-extrema/a-saddle-point-is-a-maximum-in-one-direction-and-a-minimum-in

### 2. Which conditions guarantee a local maximum at a critical point?

A. $f_{xx} < 0$
B. $f_{xx} > 0$
C. $D = f_{xx} f_{yy} - f_{xy}^2 > 0$
D. $f_x = 0$ and $f_y = 0$ (first-order conditions)

**Answer:** A. $f_{xx} < 0$; C. $D = f_{xx} f_{yy} - f_{xy}^2 > 0$; D. $f_x = 0$ and $f_y = 0$ (first-order conditions)

**Why:** All three are needed: zero gradient (it's a critical point), positive determinant (not a saddle), and negative $f_{xx}$ (concave, so it's a max, not a min).

Page: https://tryals.app/practice/business-i/unconstrained-multivariable-extrema/which-conditions-guarantee-a-local-maximum-at-a-critical-point

### 3. A saddle point is a critical point where the function is a maximum in one direction and a minimum in another.

**Answer:** True

**Why:** The mountain-pass analogy: highest point on the ridge, lowest point in the valley, a critical point that is neither max nor min overall.

Page: https://tryals.app/practice/business-i/unconstrained-multivariable-extrema/a-saddle-point-is-a-critical-point-where-the-function-is-a-maximum-in

### 4. Complete the Hessian test.

**Answer:** If $D > 0$ and $f_{xx} < 0$, the critical point is a local **maximum**. If $D < 0$, it is a **saddle** point.

**Why:** Positive $D$ with negative curvature = max; negative $D$ = saddle. The test is mechanical once you compute the three second derivatives.

Page: https://tryals.app/practice/business-i/unconstrained-multivariable-extrema/complete-the-hessian-test

### 5. Order the steps of finding and classifying a multivariable extremum.

**Answer:**

1. Compute $f_x$ and $f_y$
2. Set $f_x = 0$ and $f_y = 0$; solve for critical points
3. Compute $f_{xx}$, $f_{yy}$, $f_{xy}$ at each critical point
4. Compute $D = f_{xx} f_{yy} - f_{xy}^2$
5. Classify: sign of $D$ and $f_{xx}$ determine max/min/saddle

**Why:** Differentiate, solve, differentiate again, compute the determinant, classify, the complete recipe.

Page: https://tryals.app/practice/business-i/unconstrained-multivariable-extrema/order-the-steps-of-finding-and-classifying-a-multivariable-extremum

### 6. Match each concept to its definition.

**Answer:**

- Critical point → Where all first partial derivatives are zero
- Hessian → Matrix of second partial derivatives
- Saddle point → Neither max nor min, up in one direction, down in another
- Concavity → Surface curves downward, ensures a maximum

**Why:** Find candidates (FOC), test them (Hessian), interpret the geometry (saddle vs extremum).

Page: https://tryals.app/practice/business-i/unconstrained-multivariable-extrema/match-each-concept-to-its-definition

### 7. For $f(x,y) = x^2 - y^2$, the origin is a critical point. $f_{xx} = 2$, $f_{yy} = -2$, $f_{xy} = 0$. The determinant is:

A. $D = 0$: inconclusive
B. $D = -4$: saddle point
C. $D = 4$: local minimum
D. $D = -4$: local maximum

**Answer:** B. $D = -4$: saddle point

**Why:** $D = 2 \times (-2) - 0 = -4 < 0$: saddle point. The function rises along the $x$-axis and falls along the $y$-axis, a classic hyperbolic paraboloid.

Page: https://tryals.app/practice/business-i/unconstrained-multivariable-extrema/for-f-x-y-x-y-the-origin-is-a-critical-point-fxx-2-fyy