Question

let $f(x,y)=3x^{2}+y^{2}-2x-6y + 4$.
(a) find the minimum value of $f(x,y)$ when $y = 2$.
(b) explain why the answer to part (a) is not the minimum value of the function $f(x,y)$.

(a) the minimum value of $f(x,2)$ is \\(\square\\).
(simplify your answer.)

Explanation:

Step1: Substitute y = 2 into f(x,y)

Substitute \( y = 2 \) into \( f(x,y)=3x^{2}+y^{2}-2x - 6y + 4 \). We get \( f(x,2)=3x^{2}+2^{2}-2x-6\times2 + 4 \). Simplify the constant terms: \( 4-12 + 4=-4 \), so \( f(x,2)=3x^{2}-2x-4 \).

Step2: Find the minimum of the quadratic function

For a quadratic function \( ax^{2}+bx + c \) (here \( a = 3 \), \( b=-2 \), \( c = - 4 \)), the x - coordinate of the vertex (which gives the minimum since \( a>0 \)) is \( x=-\frac{b}{2a} \). Substitute \( a = 3 \) and \( b=-2 \) into the formula: \( x =-\frac{-2}{2\times3}=\frac{2}{6}=\frac{1}{3} \).

Step3: Substitute x = 1/3 into f(x,2)

Substitute \( x=\frac{1}{3} \) into \( f(x,2)=3x^{2}-2x - 4 \). First, calculate \( 3x^{2}=3\times(\frac{1}{3})^{2}=3\times\frac{1}{9}=\frac{1}{3} \), \( -2x=-2\times\frac{1}{3}=-\frac{2}{3} \). Then \( f(\frac{1}{3},2)=\frac{1}{3}-\frac{2}{3}-4=\frac{1 - 2}{3}-4=-\frac{1}{3}-4=-\frac{1 + 12}{3}=-\frac{13}{3} \).

Answer:

\(-\frac{13}{3}\)