Classify the stationary points of
Angles in Circles
Angles within a circle are feasible to create with the help of different properties of the circle such as radii, tangents, and chords. The radius is the distance from the center of the circle to the circumference of the circle. A tangent is a line made perpendicular to the radius through its endpoint placed on the circle as well as the line drawn at right angles to a tangent across the point of contact when the circle passes through the center of the circle. The chord is a line segment with its endpoints on the circle. A secant line or secant is the infinite extension of the chord.
Arcs in Circles
A circular arc is the arc of a circle formed by two distinct points. It is a section or segment of the circumference of a circle. A straight line passing through the center connecting the two distinct ends of the arc is termed a semi-circular arc.
Can you help with question 3?
![SECTION 3.3/ EQUALITY CONSTRAINTS: THE LAGRANGE PROBLEM
115
ROB
MS FOR SECTION 3 2
SM 1. The function
f(x1, x2, x3) = x} + x3 + 3x3
Extx + Ex1x7+ 7x]x –
defined on R has ohly one stationary point. Show that it is a local minimum point.
2. (a) Let f be defined for all (x, y) by ƒ (x, y) = x+y³-3xy. Show that (0, 0) and (1, 1) are the
only stationary points, and compute the quadratic form in (9) for ƒ at the stationary points.
(b) Check the definiteness of the quadratic form at the stationary points.
(c) Classify the stationary points by using (1)(3).
SM 3. Classify the stationary points of
(a) f(x, y, z) = x² +x?y + y²z+z? – 4z
(b) f(x1, x2, x3, X4) = 20x2 + 48x3 + 6x4 + 8x,.x2 – 4x} – 12x3 – x; – 4x3
HARDER PROBLEMS
4. Suppose f(x, y) has only one stationary point (x*, y*) which is a local minimum point. Is
(x*, y*) necessarily a global minimum point? It may be surprising that the answer is no. Prove
this by examining the function defined for all (x, y) by f (x, y) = (1 +y)³x²+y². (Hint: Look
at f(x, –2) as x → .)
3.3 Equality Constraints: The Lagrange Problem
A general optimization problem with equality constraints is of the form
max (min) f (x1, ..., Xn) subject to
1q = ("x'· · 'Ix)18
(u > w)
(1)
"q = ("x' · 'x) 48
where the b; are constants. We assume that m < n because otherwise there are usually no
degrees of freedom.
In vector formulation, the problem is
12:5
4/18](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3143ece0-b2b2-417b-825f-ab4e468fd03e%2F22fa6658-6329-47a2-bd42-e78bc6be57f4%2Fguwp7ja_processed.jpeg&w=3840&q=75)
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