Find the absolute maximum and minimum values off on the set D. f\left(x,

vomiderawo 2021-11-18 Answered
Find the absolute maximum and minimum values off on the set D.
\(\displaystyle{f}{\left({x},{y}\right)}={x}^{{{2}}}+{y}^{{{2}}}-{2}{x}\), D is the closed triangular region with vertices \(\displaystyle{\left({2},{0}\right)}\), \(\displaystyle{\left({0},{2}\right)}\), and \(\displaystyle{\left({0},-{2}\right)}\)

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Expert Answer

hotcoal3z
Answered 2021-11-19 Author has 1415 answers
Step 1
Start with the given function. Skelch the domain (a trangle), and label the three vertices of the triangle For the fust step, find the values of any cntical points within the domain
\(\displaystyle{f}{\left({x},{y}\right)}={x}^{{{2}}}+{y}^{{{2}}}-{2}{x}\)
\(\displaystyle{A}={\left({0},{2}\right)}\)
\(\displaystyle{B}={\left({2},{0}\right)}\)
\(\displaystyle{C}={\left({0},-{2}\right)}\)
Step 2
Find all the critical points by taking the first derivative of the function with respect to both x and y, and then solving for \(\displaystyle{f}_{{{x}}}={0}\) and \(\displaystyle{f}_{{{y}}}={0}\).
\(\displaystyle{f}_{{{x}}}={2}{x}-{2}={0}\rightarrow{x}={1}\)
\(\displaystyle{f}_{{{y}}}={2}{y}={0}\rightarrow{y}={0}\)
Value at critical point: \(\displaystyle{f}{\left({1},{0}\right)}=-{1}\)
This gives us a single critical point at \(\displaystyle{f}{\left({1},{0}\right)}\) with a value of 1, which is within the domain.
Absolute minimum at critical point: \(\displaystyle{f}{\left({1},{0}\right)}=-{1}\)
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Opeance1951
Answered 2021-11-20 Author has 601 answers

Step 1
For the second step, we need to calculate the minima and maxima values along each boundary of the domain -i.e., along the three edges of the triangle.
Start by solving for the equation of the line segment that connects each point.
\(\displaystyle\overline{{{A}{B}}}:{y}={2}-{x},{0}\le{x}\le{2}\)
\(\displaystyle\overline{{{B}{C}}}:{y}={x}-{2},{0}\le{x}\le{2}\)
\(\displaystyle\overline{{{A}{C}}}:{x}={0},-{2}\le{y}\le{2}\)
Step 2
Solve for the values of the function along the boundary \(\displaystyle\overline{{{A}{B}}}\) in terms of x by substituting \(\displaystyle{y}={2}-{x}\) in the original equation and simplifying
Since the result i a quadratic, find the x coordinate of the inflection point along this line by taking the first derivative and solving for 0, then find the corresponding y coordinate.
\(\displaystyle{f}{\left({x},{2}-{x}\right)}={x}^{{{2}}}+{\left({2}-{x}\right)}^{{{2}}}-{2}{x}={2}{x}^{{{2}}}-{6}{x}+{4}\)
\(\displaystyle{f}'{\left({x},{2}-{x}\right)}={4}{x}-{6}={0}\rightarrow{x}={\frac{{{3}}}{{{2}}}},{y}={\frac{{{1}}}{{{2}}}}\)
Step 3
Calculate the values of \(\displaystyle{f}{\left({x},{y}\right)}\) at the two end points of \(\displaystyle\overline{{{A}{B}}}\) and at its inflection point.
\(\displaystyle{f}{\left({0},{2}\right)}={4}\)
\(\displaystyle{f}{\left({\frac{{{3}}}{{{2}}}},{\frac{{{1}}}{{{2}}}}\right)}=-{\frac{{{1}}}{{{2}}}}\)
\(\displaystyle{f}{\left({2},{0}\right)}={0}\)
Step 4
Next solve for the values of the function along the boundary \(\displaystyle\overline{{{B}{C}}}\) by subisifuling \(\displaystyle{y}={x}-{2}\) and simplifying (This results in the same equation as for \(\displaystyle\overline{{{A}{B}}}\) above, since \(\displaystyle{\left({x}-{2}\right)}^{{{2}}}={\left({2}-{x}\right)}^{{{2}}}\).) Then find the inflection point as described above.
\(\displaystyle={\left({x},{x}-{2}\right)}={x}^{{{2}}}+{\left({x}-{2}\right)}^{{{2}}}-{2}{x}={2}{x}^{{{2}}}-{6}{x}+{4}\)
\(\displaystyle{f}'{\left({x},{x}-{2}\right)}={4}{x}-{6}={0}\rightarrow{x}={\frac{{{3}}}{{{2}}}},{y}=-{\frac{{{1}}}{{{2}}}}\)
Step 5
Calculate the values of \(\displaystyle{f}{\left({x},{y}\right)}\) at the two end points of \(\displaystyle\overline{{{B}{C}}}\) and at its inflection point.
\(\displaystyle{f}{\left({0},-{2}\right)}={4}\)
\(\displaystyle{f}{\left({\frac{{{3}}}{{{2}}}},-{\frac{{{1}}}{{{2}}}}\right)}=-{\frac{{{1}}}{{{2}}}}\)
\(\displaystyle{f}{\left({2},{0}\right)}={0}\) (same as above)
Step 6
Finally, solve for the equation of the line segment \(\displaystyle\overline{{{A}{C}}}\). To do this, substitute \(\displaystyle{y}={0}\) in the original equation. This has an inflection point (a minimum value) at \(\displaystyle{\left({0},{0}\right)}\).
\(\displaystyle{f}{\left({0},{y}\right)}={y}^{{{2}}}\)
Step 7
Determine the values of function at the two end points along the boundary \(\displaystyle\overline{{{A}{C}}}\) and at this inflection point. Since we've already calculated the values at the end points \(\displaystyle{\left({0},{2}\right)}\) and \(\displaystyle{\left({0},-{2}\right)}\) above, there's no need to do so again.
\(\displaystyle{f}{\left({0},{0}\right)}={0}\) Step 8
Find the absolute maximums and minimums on the domain by comparing the values of the critical point, the end points of each boundary segment, and each inflection point The minimum is the smallest value, and the maximum is the largest.
Absolute maximums at \(\displaystyle{f}{\left({0},{2}\right)}={f}{\left({0},-{2}\right)}={4}\)

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