Chapter 13: Problem 8
What is the procedure for locating absolute maximum and minimum values on a closed bounded domain?
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Chapter 13: Problem 8
What is the procedure for locating absolute maximum and minimum values on a closed bounded domain?
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A function of one variable has the property that a local maximum (or minimum) occurring at the only critical point is also the absolute maximum (or minimum) (for example, \(f(x)=x^{2}\) ). Does the same result hold for a function of two variables? Show that the following functions have the property that they have a single local maximum (or minimum), occurring at the only critical point, but that the local maximum (or minimum) is not an absolute maximum (or minimum) on \(\mathbb{R}^{2}\). a. \(f(x, y)=3 x e^{y}-x^{3}-e^{3 y}\) b. \(f(x, y)=\left(2 y^{2}-y^{4}\right)\left(e^{x}+\frac{1}{1+x^{2}}\right)-\frac{1}{1+x^{2}}\) This property has the following interpretation. Suppose that a surface has a single local minimum that is not the absolute minimum. Then water can be poured into the basin around the local minimum and the surface never overflows, even though there are points on the surface below the local minimum.
Use the method of your choice to ate the following limits. $$\lim _{(x, y) \rightarrow(0,0)} \frac{|x-y|}{|x+y|}$$
Use the Second Derivative Test to prove that if \((a, b)\) is a critical point
of \(f\) at which \(f_{x}(a, b)=f_{y}(a, b)=0\) and \(f_{x x}(a, b)<0
Prove that for the plane described by \(f(x, y)=A x+B y,\) where \(A\) and \(B\) are nonzero constants, the gradient is constant (independent of \((x, y)\) ). Interpret this result.
Let \(R\) be a closed bounded set in \(\mathbb{R}^{2}\) and let \(f(x, y)=a x+b y+c,\) where \(a, b,\) and \(c\) are real numbers, with \(a\) and \(b\) not both zero. Give a geometrical argument explaining why the absolute maximum and minimum values of \(f\) over \(R\) occur on the boundaries of \(R\)
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