Find the Minimum Value of f(x,y) = x² + y² + 6x + 12 | Maxima & Minima | Example 1 | Applied Calculus

Maxima & Minima · Worked Example 1

Find the minimum value of \( f(x,y) = x^2 + y^2 + 6x + 12 \)

Step 1 — First Order Partial Derivatives

\[ f_x = \frac{\partial f}{\partial x} = 2x + 6 \qquad\qquad f_y = \frac{\partial f}{\partial y} = 2y \]

Step 2 — Stationary Point

Setting both partial derivatives to zero:

\[ f_x = 0 \implies 2x + 6 = 0 \implies x = -3 \] \[ f_y = 0 \implies 2y = 0 \implies y = 0 \]

So the stationary point is \((-3, 0)\).

Step 3 — Second Order Partial Derivatives

\[ A = f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2 \qquad B = f_{xy} = \frac{\partial^2 f}{\partial y\,\partial x} = 0 \qquad C = f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2 \]

Step 4 — Nature of the Stationary Point

Test	At (-3, 0)
A = 2	2 > 0
B = 0	0
C = 2	2
\(AC - B^2\)	4 − 0 = 4 > 0
Minimum

Step 5 — Minimum Value

\[ f(-3,0) = (-3)^2 + 0^2 + 6(-3) + 12 = 9 - 18 + 12 = 3 \]

Final Answer

Minimum value of \(f(x,y)\) at \((-3,0)\) is \(3\).

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3D View — The Paraboloid

\(f(x,y)=x^2+y^2+6x+12\) is a bowl-shaped paraboloid. The lowest point of the bowl (the red dot below) sits exactly at \((-3,0,3)\) — the minimum we found above. Drag to rotate.

🖱️ Click and drag inside the box to rotate the surface

[JAN-22-R-21]

Watch it worked out Example 1

Maxima & Minima Series · Worked Examples for Board / Competitive Exams

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