Question

Find all points​ (x,y) where​ f(x,y) has a possible relative maximum or minimum.​ Then, use the​ second-derivative test to​ determine, if​ possible, the nature of​ f(x,y) at each of these points. If the​ second-derivative test is​ inconclusive, so state. ​f(x,y)=5x2+10xy+6y2+10x+12y+9

Answer 1

The points where
`\( f(x,y) \)` has possible relative maximum or minimum are at the critical points. To find these points, we'll take the partial derivatives of
`\( f(x,y) \)` with respect to
`\( x \)` and
`\( y \)`, set them equal to zero, and solve for
`\( x \)` and
`\( y \)`. Then, we'll utilize the second-derivative test to determine the nature of
`\( f(x,y) \)` at each of these critical points.

Explanation:

Certainly! Let's break down the process mathematically:

Given the function
`\( f(x,y) = 5x^2 + 10xy + 6y^2 + 10x + 12y + 9 \)`, to find critical points, we first calculate the partial derivatives:

`\((\partial f)/(\partial x) = 10x + 10y + 10\)`

`\((\partial f)/(\partial y) = 10x + 12y + 12\)`

Setting these derivatives equal to zero to find critical points:

`\(10x + 10y + 10 = 0\)`

`\(10x + 12y + 12 = 0\)`

Solving these simultaneous equations gives the critical point
`\( (x, y) = (-1, -1) \)`.

For the second-derivative test, we compute the second partial derivatives:

`\((\partial^2 f)/(\partial x^2) = 10\)`

`\((\partial^2 f)/(\partial y^2) = 12\)`

`\((\partial^2 f)/(\partial x \partial y) = 10\)`

Evaluating these second partial derivatives at the critical point
`\( (-1, -1) \)`:

`\((\partial^2 f)/(\partial x^2) = 10\)`

`\((\partial^2 f)/(\partial y^2) = 12\)`

`\((\partial^2 f)/(\partial x \partial y) = 10\)`

Constructing the determinant
`\( D = (\partial^2 f)/(\partial x^2) \cdot (\partial^2 f)/(\partial y^2) - \left((\partial^2 f)/(\partial x \partial y)\right)^2 \)` and evaluating it at the critical point:

`\(D = (10)(12) - (10)^2 = 20\)`

Since
`\( D > 0 \)`and
`\( (\partial^2 f)/(\partial x^2) > 0 \)` at the critical point
`\( (-1, -1) \)`, the second-derivative test confirms that
`\( f(x,y) \)`has a relative minimum at
`\( (-1, -1) \)`.