# Find all relative extreme of 3x^2 + 4y^2 \ over \ the \ rectangle \ 0 \leq x \leq 2, 0 \leq y...

## Question:

Find all relative extreme of {eq}3x^2 + 4y^2 \ over \ the \ rectangle \ 0 \leq x \leq 2, 0 \leq y \leq 1 {/eq}

## Critical Points:

To determine the critical points of a function of two variables, it is necessary to calculate the first derivatives.

The points for which both derivatives are equal to zero, are the points for which a critical point of the function is reached.

## Answer and Explanation:

Denoting the function by, {eq}f\left( {x,y} \right) = 3{x^2} + 4{y^2} {/eq}, calculating the first partial derivatives:

{eq}{f_x}\left( {x,y} \right) = 6x = 0\\ {f_y}\left( {x,y} \right) = 8y = 0 {/eq}

So, we have the point, {eq}x = 0,y = 0 \to \left( {0,0} \right). {/eq}

Taking into account the function is always positive (sum of squares), at the origin, a local minimum is reached.

#### Learn more about this topic: Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
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