Question

What are the values and types of the critical points, if any, of `f(x,y)=xy(1-8x-7y)`?

Answer 1

# {: ("Critical Point","Conclusion"), ((0,0,0),"saddle"), ((0,1/7,0),"saddle"), ((1/8,0,0),"saddle"), ((1/24,1/21,1/1512),"max") :} #

Explanation:

The theory to identify the extrema of `z=f(x,y)` is:

1. Solve simultaneously the critical equations

   `(partial f) / (partial x) =(partial f) / (partial y) =0 \ \ \` (ie `f_x=f_y=0`)

2. Evaluate `f_(x x), f_(yy) and f_(xy) (=f_(yx))` at each of these critical points. Hence evaluate `Delta=f_(x x)f_(yy)-f_(xy)^2` at each of these points
3. Determine the nature of the extrema;

   `{: (Delta>0, "There is minimum if " f_(x x)<0),(, "and a maximum if " f_(yy)>0), (Delta<0, "there is a saddle point"), (Delta=0, "Further analysis is necessary") :}`

So we have:

`f(x,y) = xy(1-8x-7y)`
`" " = xy -8x^2y-7xy^2`

Let us find the first partial derivatives:

`(partial f) / (partial x) = y-16xy-7y^2`
`(partial f) / (partial y) = x-8x^2-14xy`

So our critical equations are:

`y-16xy-7y^2 = 0`
`x-8x^2-14xy = 0`

From the First equation we have:

`y(1-16x-7y) = 0=> y=0 or 1-16x-7y = 0`

From the Second equation we have:

`x(1-8x-14y) = 0 => x=0 or 1-8x-14y`

`x=0 => 1-7y=0 => y=1/7`
`y=0 => 1-8x=0 => x=1/8`

And simultaneously (solution not shown) we get the solution

`x=1/24; y=1/21`

And so we have four critical points with coordinates;

`(0,0,0), (0,1/7,0), (1/8,0,0), (1/24,1/21,1/1512)`

So, now let us look at the second partial derivatives so that we can determine the nature of the critical points:

`\ \ \ (partial^2f) / (partial x^2) = -16y`
`\ \ \ (partial^2f) / (partial y^2) = -14x`
`(partial^2f) / (partial x partial y) = 1-16x-14y \ \ \ \ (=(partial^2f) / (partial y partial x))`

And we must calculate:

`Delta=(partial^2f) / (partial x^2) (partial^2f) / (partial y^2) - ((partial^2f) / (partial x partial y))^2`

at each critical point. The second partial derivative values, `Delta`, and conclusion are as follows:

# {: ("Critical Point",(partial^2f) / (partial x^2),(partial^2f) / (partial y^2),(partial^2f) / (partial x partial y),Delta,"Conclusion"), ((0,0,0),0,0,1,lt 0,"saddle"), ((0,1/7,0),-16/7,0,-1,lt 0,"saddle"), ((1/8,0,0),0,-7/4,-1,lt 0,"saddle"), ((1/24,1/21,1/1512),-16/21,-7/12,-1/3,gt 0,f_(x x)<0 => "max") :} #

We can see these critical points if we look at a 3D plot. The undulations in the surface are really subtle so this is a highly zoomed plot: