Absolute maximum and minimum calculator two variables

Q: How do you find the absolute max and min on a graphing calculator?

Press 2nd, then CALC (above the TRACE key). Choose 4: maximum or 3: minimum. The calculator will ask you for a left bound, a right bound, and a guess for the maximum or minimum. You can enter these by using your left and right arrows to move the cursor to a reasonable x-value, then pressing ENTER.

The above calculator is an online tool which shows output for the given input. This calculator, which makes calculations very simple and interesting. If an input is given then it can easily show the result for the given number.

From your calculation, you have shown that there is no critical point in the interior of the half-disk, but the left endpoint of the semicircular boundary is one. The index of that point is

$$ D \ \ = \ \ f_{xx}·f_{yy} \ - \ (f_{xy})^2 \ \ = \ \ 2·(-2) \ - \ 0^2 \ = \ -4 \ \ < \ \ 0 \ \ , $$

which identifies $ \ (-1 \ , \ 0 ) \ $ as a saddle-point. This makes sense since the function is "concave upward" in the $ \ x-$ direction about $ \ x \ = \ -1 \ $ , due to the $ \ x^2 \ + \ 2x \ = \ (x + 1)^2 \ - \ 1 \ $ terms, but is "concave downward" in the $ \ y-$ direction because of the $ \ -y^2 \ $ term. As we will need it for comparisons later, we compute that $ f(-1,0) \ = \ (-1)^2 \ + \ 2(-1) \ - \ 0^2 \ = \ -1 \ \ . $

The lower boundary of the region is the $ \ x-$ axis over the interval $ \ [ \ -1 \ , \ 1 \ ] \ $ . Inserting $ \ y \ = \ 0 \ $ into the function expression gives $ \ f(x,0) \ = \ (x + 1)^2 \ - \ 1 \ \ . $ We have already found $ \ f(-1,0) \ $ to be the minimum on this line segment; the right endpoint has the value $ f(1,0) \ = \ (1+1)^2 \ - \ 1 \ = \ 3 \ \ . $

It remains to examine the upper semicircular boundary. There are a number of ways we might set this up, but to avoid having to work with a square-root or trigonometric identities, we will use the circle equation to write $ \ y^2 \ = \ 1 \ - \ x^2 \ $ and express our function on this section of the boundary as

$$ \ \phi(x) \ = \ (x + 1)^2 \ - \ 1 \ - \ (1 - x^2) \ \ = \ \ (x + 1)^2 \ + \ x^2 \ - \ 2 \ \ . $$

[This gives us the correct values for the function at the endpoints of the region:

$$ \phi(-1) \ \ = \ \ ([-1] + 1)^2 \ + \ (-1)^2 \ - \ 2 \ \ = \ -1 \ \ \ , \ \ \ \phi(1) \ \ = \ \ (1 + 1)^2 \ + \ 1^2 \ - \ 2 \ \ = \ \ 3 \ \ . \ ] $$

We find $ \ \phi'(x) \ = \ 2·(x + 1) \ + \ 2x \ \ = \ \ 0 \ \ \ \Rightarrow \ \ \ 4x \ \ = \ -2 \ \ \ \Rightarrow \ \ \ x \ = \ -\frac{1}{2} \ \ , $ for which $ \phi \left( -\frac{1}{2} \right) \ \ = \ \ \left( \ \left[-\frac{1}{2} \right] + 1 \ \right)^2 \ + \ \left(-\frac{1}{2} \right)^2 \ - \ 2 \ \ = \ \ \frac{1}{4} \ + \ \frac{1}{4} \ - \ 2 \ \ = \ -\frac{3}{2} \ \ . $

Hence, we observe that the absolute maximum of the function on this closed half-disc is $ f(1,0) \ = \ 3 \ $ and the absolute minimum is $ f \left( -\frac{1}{2} \ , \ \frac{\sqrt{3}}{2} \right) \ = \ -\frac{3}{2} \ \ . $ The saddle point at $ \ (-1 \ , \ 0 ) \ $ has no special role in this region.

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Absolute Maximum/Minimum Values of Multivariable Functions - Part 1 of 2
To find absolute max/min values of a continuous function g on a closed bounded set D:

Evaluate f at the critical points of f in D.
Find the extreme values of f on the boundary of D.
Pick the largest and smallest.

Example:
Find the absolute maximum and minimum of:
f(x,y) = 3 + xy - x - 2y; D is the closed triangular region with vertices (1,0), (5,0), (1,4).

Absolute Maximum/Minimum Values of Multivariable Functions - Part 2 of 2

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Find the critical points.

Find the first derivative.

By the Sum Rule, the derivative of with respect to is .

Differentiate using the Power Rule which states that is where .

Since is constant with respect to , the derivative of with respect to is .

The first derivative of with respect to is .

Set the first derivative equal to then solve the equation .

Set the first derivative equal to .

Divide each term in by and simplify.

Cancel the common factor of .

Cancel the common factor.

Find the values where the derivative is undefined.

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Evaluate at each value where the derivative is or undefined.

Raising to any positive power yields .

How do you find the absolute maximum and minimum?

Finding the Absolute Extrema.

Find all critical numbers of f within the interval [a, b]. ... .

Plug in each critical number from step 1 into the function f(x)..

Plug in the endpoints, a and b, into the function f(x)..

The largest value is the absolute maximum, and the smallest value is the absolute minimum..

How do you find the absolute max and min on a graphing calculator?

Press 2nd, then CALC (above the TRACE key). Choose 4: maximum or 3: minimum. The calculator will ask you for a left bound, a right bound, and a guess for the maximum or minimum. You can enter these by using your left and right arrows to move the cursor to a reasonable x-value, then pressing ENTER.