WORKED EXAMPLES2 · Engineering Mathematics 2 Solved Problems
Finding the Minimum of a Two‑Variable Function
Two full walkthroughs of stationary points and the second‑derivative test, applied to \(f(x,y) = x^2 + y^2 + 6x + 12\) and \(f(x,y) = x^2 - xy + y^2 - 2x + y\).
- #PartialDerivatives
- #StationaryPoints
- #SecondDerivativeTest
- #CalculusOfTwoVariables
Watch first — Example 2
Locating local maxima, minima, and saddle points for a function of two variables — a companion explainer on YouTube.
Before the algebra, the idea is simple: a smooth surface \(z = f(x,y)\) flattens out at its peaks, valleys, and saddles. Set both partial derivatives to zero to find those flat points, then use a second‑derivative test to tell which kind each one is.
📐 The worked examples
Two standard board questions, each solved step by step.
Find the minimum value of \(f(x,y) = x^2 + y^2 + 6x + 12\)JAN‑22‑R‑21
Step 1 — First partial derivatives
Step 2 — Locate the stationary point
The stationary point is \((-3,\,0)\).
Surface of \(f(x,y) = x^2+y^2+6x+12\) — the bowl bottoms out at the minimum, \((-3,\,0,\,3)\).
Step 3 — Second partial derivatives
Step 4 — Classify the point
| At \((-3,\,0)\) | |
|---|---|
| A = 2 | 2 > 0 |
| B = 0 | 0 |
| C = 2 | 2 |
| \(AC - B^2\) | 4 − 0 = 4 > 0 |
| Minimum |
Since \(AC - B^2 > 0\) and \(A > 0\), \((-3,\,0)\) is a minimum point.
Step 5 — Evaluate the minimum value
Find the minimum value of \(f(x,y) = x^2 - xy + y^2 - 2x + y\)N/D‑22‑R‑21
Step 1 — First partial derivatives
Step 2 — Locate the stationary point
Solving the two equations together gives \(x = 1\), \(y = 0\), so the stationary point is \((1,\,0)\).
Surface of \(f(x,y) = x^2-xy+y^2-2x+y\) — the tilted bowl bottoms out at the minimum, \((1,\,0,\,-1)\).
Step 3 — Second partial derivatives
Step 4 — Classify the point
| At \((1,\,0)\) | |
|---|---|
| r = 2 | 2 > 0 |
| s = −1 | −1 |
| t = 2 | 2 |
| \(rt - s^2\) | 4 − 1 = 3 > 0 |
| Minimum |
Since \(rt - s^2 > 0\) and \(r > 0\), \((1,\,0)\) is a minimum point.
Step 5 — Evaluate the minimum value
💡 Why this matters
Three things worth keeping from this example.
- A stationary point only tells you where the surface is flat — the second‑derivative test tells you what kind of flat it is.
- \(AC - B^2 > 0\) rules out a saddle; the sign of \(A\) (or \(C\)) then decides maximum versus minimum.
- Completing the square confirms it instantly here: \(f(x,y) = (x+3)^2 + y^2 + 3\), which is never below 3.
No comments:
Post a Comment