Absolute Extreme Values

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Problem set 9, question 2 (finding tangents to level curves)?

Part A asks you to find the gradient of a function. This follows textbook and class examples closely.

For Part B, the gradient is a vector in the xy plane that is perpendicular to the level curve through the given point. So to find a tangent to the level curve, just find a vector whose dot product with the gradient is 0. One elegant way to do this in 2 dimensions is to exchange the components of the gradient and negate one of them.

Part C requires more knowledge of the level curve. The question gives you the constant value for the level curve, so setting z equal to that value gives an implicit equation for the level curve, which happens to be the equation for a circle. The simplest parametric form for circle is r(t) = ⟨ a cos t, a sin t ⟩ where a is the circle’s radius. This gives you a parametric equation for the circle, and its derivative gives you an equation for tangents to the circle. Solve r(t) = (√2, -√2) to find the t value corresponding to the point the question asks about, and plug that t value into the derivative to get the tangent:

Finally, you can test to see if the tangent vectors from parts B and C are equal by checking whether they are scalar multiples of each other.

Absolute Extreme Values

Based on “Absolute Maxima and Minima” in section 3.7.

Key Ideas

To find absolute minima and maxima on a closed, bounded, region, you need to check the function’s values on the boundaries of the region and at critical points inside it.

This gives rise to the following procedure for finding absolute extrema:

1. Find critical points
2. Evaluate the function at the critical points
3. Determine the boundary and any extremes along it (by finding critical points of the function restricted to the boundary)
4. The absolute extremes are the smallest and largest function values occurring at the critical points or on the boundary.

The Extreme Value Theorem provides the theoretical basis for this procedure. It says that a function that is continuous on a closed and bounded domain must have absolute extreme values.

Questions

How does whether the domain is open or closed affect whether extremes are local or absolute? You know they’re absolute within the domain if the domain is closed (and bounded). On the other hand, an open domain might not have absolute extreme values at all.

Example

Find the absolute minimum and maximum values of z = 2x² + 3y² - 4x - 6y + 5 over the triangle in the xy plane with vertices (0,0), (3,0), and (0,3) (including those vertices and the sides of the triangle).

We started by finding critical values for the function. There’s one critical point, which is inside the triangle, so it goes on the list of places to check the function value:

Then we checked the sides of the triangle. Two of them correspond to one or the other of x or y being 0, which allows us to simplify the function to a function of just 1 variable. We used ordinary single-variable derivatives to find points along these edges where there might be extremes. This added 2 more points to the list of places to check the function’s value:

We still have the third side of the triangle (between points (3,0) and (0,3)) to check, and each of the vertices of the triangle. We’ll finish these tomorrow.

Next

Finish the absolute extreme value problem from above.

Start talking about constrained optimization problems and the method of Lagrange multipliers.

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