Applications of Derivatives
Finding Extrema of Multivariable Functions
Grade 12
Question:
<p>Let \(f(x,y) = x^2 - 2xy - 3y^2 - 6x - 2y\), where \(x, y \in \mathbb{R}\), then</p>
<p>(a) \(f(x,y) \leq -11\)</p>
<p>(b) \(f(x,y) \leq -10\)</p>
<p>(c) \(f(x,y) \geq -11\)</p>
<p>(d) \(f(x,y) \geq -12\)</p>
Step-by-Step Solution
Key Concept: Treat the expression as a quadratic in one variable and use the discriminant condition to find bounds on the function value.
<p><strong>Solution:</strong> Let $z = x^2 - 2xy - 3y^2 - 6x - 2y$</p><p>For x to be real, treating this as a quadratic in x:</p><p>$x^2 - 2xy - 3y^2 - 6x - 2y - z = 0$</p><p>Discriminant $\Delta \geq 0$:</p><p>$4(y+3)^2 - 4(3y^2 - 2y - z) \geq 0$</p><p>$y^2 + 9 + 6y - 3y^2 + 2y + z \geq 0$</p><p>$-2y^2 - 4y + 9 + z \geq 0$</p><p>$z \geq 2(y^2 + 2y - 1) - 11 = 2(y+1)^2 - 11$</p><p>Minimum value: $z \geq -11$ when $y = -1$</p><p>Also, $z \leq -11$ has a maximum at $z = -11$.</p><p>∴ Answers are (a), (c) and (d) are correct.</p>
Correct Answer: A,C