Let $f ( x , y ) = x ^ { 2 } + y ^ { 2 }$ with constraint $3 x + y = 3$ . Then f has a relative minimum at

A) $\left( \frac { 9 } { 10 } , \frac { 3 } { 10 } \right)$
B) $\left( - \frac { 9 } { 10 } , \frac { 3 } { 10 } \right)$
C) $\left( \frac { 9 } { 10 } , - \frac { 3 } { 10 } \right)$
D) $\left( \frac { 3 } { 10 } , \frac { 9 } { 10 } \right)$
E) $\left( \frac { 3 } { 10 } , - \frac { 9 } { 10 } \right)$

Question

Let $f ( x , y ) = x ^ { 2 } + y ^ { 2 }$ with constraint $3 x + y = 3$ . Then f has a relative minimum at

A) $\left( \frac { 9 } { 10 } , \frac { 3 } { 10 } \right)$
B) $\left( - \frac { 9 } { 10 } , \frac { 3 } { 10 } \right)$
C) $\left( \frac { 9 } { 10 } , - \frac { 3 } { 10 } \right)$
D) $\left( \frac { 3 } { 10 } , \frac { 9 } { 10 } \right)$
E) $\left( \frac { 3 } { 10 } , - \frac { 9 } { 10 } \right)$

Quizplus · Accepted Answer

To find the relative minimum of $f(x, y) = x^2 + y^2$ subject to the constraint $3x + y = 3$, we can use the method of Lagrange multipliers or solve the constraint for one variable and substitute back into $f$. Solving the constraint for $y$, we get $y = 3 - 3x$. Substituting into $f$, we get $f(x, 3-3x) = x^2 + (3-3x)^2$. Differentiating this expression with respect to $x$ and setting the derivative equal to zero will give the value of $x$ that minimizes $f$. Substituting this $x$ value back into the constraint equation gives $y$, and the pair $(x, y)$ that minimizes $f$ under the given constraint is $\left( \frac { 9 } { 10 } , \frac { 3 } { 10 } \right)$.

Let f(x,y)=x2+y2f ( x , y ) = x ^ { 2 } + y ^ { 2 }f(x,y)=x2+y2

Let $f ( x , y ) = x ^ { 2 } + y ^ { 2 }$