Question

Find the absolute maximum and minimum of the function on the given domain. f (x comma y )equals 2 x squared plus 5 y squared on the closed triangular plate bounded by the lines x equals 0 comma y equals 0 comma y plus 2 x equals 2 in the first quadrant

Answer 1

The absolute maximum of
`f(x,y)` in the triangle plate is 20, and the absolute minimum is 0.

Explanation:

We have the function
`f(x,y)=2x^2+5y^2` and we want to the the absolute maximum and minimum in the bounded region determined by the lines
`y=0`,
`x=0` and
`y+2x=2`, and we will call it
`A`.

Usually, in this kind of problem, the first step is to assure the existence of the absolute maximum and minimum. The general procedure is to show that the function is continuous and the region is closed and bounded.

It is no difficult to see that the function
`f(x,y)` is continuous in the whole plane, so it is in the given region. Moreover, the region is closed and bounded. So, we can assure the existence of the absolute minimum and maximum of
`f(x,y)` in the triangle plate.

The second step is to find the critical point(s) of the function in the interior of
`A`. In order to do that we must calculate the partial derivatives:

`(\partial f)/(\partial x)(x,y) = 4x`

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

Recall that the critical points are those that makes zero both partial derivatives. So, we need to solve the system of equations

`4x=0`

`10y=0`

that has as only solution the point
`(x,y)=(0,0)`. Now, the point (0,0) is not in the interior of
`A` but in the boundary.

As the function has no critical points inside the region
`A`, we must look for the extremal values of
`f(x,y)` in the boundary of
`A`.

In order to make the following explanation simpler there is a figure attached, where the triangle plate
`A` has been drawn.

Now, we need to give ‘‘parametric’’ expressions for the boundary of the set
`A`. Notice that:

• The segment AC has the form
  `(x,y) = (x,0)` with
  `x\in[0,1]`.
• The segment AB has the form
  `(x,y) = (0,y)` with
  `y\in[0,2]`.
• The segment BC has the form
  `(x,y) = (x,2-2x)` with
  `x\in[0,1]`

The last representation was deduced from the equation of the line, because
`y+2x=2` is equivalent to
`y=2-2x`.

Now, let us analyze the behavior of
`f(x,y)` in each of the sides of the triangle. The procedure will be to find the maximal and minimum of
`f(x,y)` in each side of the triangle, and finally we will compare them and take the minimal value as the absolute minimum and the maximal as the absolute maximum.

The segment AC. In this case we have
`f(x,0) =2x^2` with
`x\in[0,1]`. So, we have reduced our two variable function to one variable function, which has a simpler behavior. Here, we can affirm that the maximal value of
`f(x,y)` in AC is obtained when
`x=1` and is
`f(1,0)=2`, and the minimal is obtained when
`x=0` and is
`f(0,0)=0`.

The segment AB. In this case we have
`f(0,y) =5y^2` with
`y\in[0,2]`. So, again, we have reduced our two variable function to one variable function, which has a simpler behavior. Here, we can affirm that the maximal value of
`f(x,y)` in AB is obtained when
`y=2` and is
`f(0,2)=20`, and the minimal is obtained when
`y=0` and is
`f(0,0)=0`.

The segment BC. In this case we have

`f(x,2-2x) =2x^2 + 5(2-2x)^2=22x^2-40x+20` with
`x\in[0,1]`.

So, we have reduced our two variable function to one variable function. In order to obtain the extremal values of
`f(x,y)` in BC, we must find the extremal values of
`22x^2-40x+20=g(x)` in the interval [0,1]. Though not so easy as the previous two, this problem is easier that the original.

So, we calculate the derivative of the function
`g(x)`:

`g'(x) = 44x-40`, which has only one zero in the interval [0,1] and it is 40/44. To obtain the minimum and maximum of
`g(x)` in [0,1] we only need to evaluate
`g(0)`,
`g(40/44)` and
`g(1)`. Then,

• 
  `g(0) = 20`

• 
  `g(40/44) \approx 1.818`

• 
  `g(1) = 2`

So, we can affirm that the maximal value of
`f(x,y)` in BC is obtained when
`x=0` and is
`f(0,2)=20`, and the minimal is obtained when
`x=40/44` and is
`f(40/44,0.181)=1.181`.

Conclusion: The extremal values of
`f(x,y)` obtained in the boundary of the triangle are:

• 
  `f(1,0)=2` and
  `f(0,0)=0`.

• 
  `f(0,2)=20` and
  `f(0,0)=0`.

• 
  `f(0,2)=20` and
  `f(40/44,0.181)=1.181`.

So, of all those values the minimum is
`f(0,0)=0` and the maximum is
`f(0,2)=20`. This means that

`\max_((x,y)\in A)\{f(x,y)\} = 20`

`\min_((x,y)\in A)\{f(x,y)\} = 0`.