Best point on a fixed curve

Mathematics · Calculus & Analysis · ages 19-20

Name ______________________ Date ____________

For f=x²+3y² with constraint g=2x+y=7, which system uses grad f=lambda grad g?
- 2x = lambda; 6y = 2lambda; 2x + y = 7
- 2x = 2lambda; 3y = lambda; 2x + y = 7
- 2x = 2lambda; 6y = lambda; 2x + y = 7
- 2x = 2lambda; 6y = lambda; x + 2y = 7
Use multipliers on f(x,y)=x+y with x²+y²=2. What is the largest value of f?
- 2
- 4
- 1
At a constrained extremum, two smooth level curves share a tangent. Both gradients are nonzero. What is their relationship?
- Their gradients are perpendicular
- Their gradients are parallel, possibly opposite
- Both gradients point along the common tangent line
A rectangle has positive side lengths and perimeter 20. Use multipliers to find its largest possible area.
Answer: ______________
Minimise f=x subject to x+y=7 and x,y>=0. What is the minimum value, including the endpoints?
Answer: ______________
Use grad f=lambda grad g with g=c. Wren says that along a smooth branch of regular optimal points, lambda is the derivative of the optimal value with respect to c. Is this true?
Circle one: True False
For real x and y with x+y=4, what is the smallest value of x²+y²?
Answer: ______________
For f=x with g=x²+y²=0, the origin is the only feasible point. Why can it be an extremum even though no lambda solves grad f=lambda grad g?
- grad g is zero, so the regular theorem does not apply
- grad g is nonzero, so lambda must be zero
- grad f is zero, so every value of lambda solves the equations
- The constraint has no feasible point, so f has no extremum
For real x,y,z with x+y+z=3, what is the smallest value of x²+y²+z²?
Answer: ______________
For real x,y,z with x+2y+2z=9, what is the smallest value of x²+y²+z²?
Answer: ______________

Answer key

For grown-ups. Fold this page away before handing over the rest.

Best point on a fixed curve W1-mt_0RPOBWZ1gs-s1

2x = 2lambda; 6y = lambda; 2x + y = 7 · The gradients are (2x,6y) and (2,1). Matching components gives 2x=2lambda and 6y=lambda, together with 2x+y=7.
2 · The candidates (1,1) and (-1,-1) give values 2 and -2. Comparing them gives the global maximum 2.
Their gradients are parallel, possibly opposite · Each gradient is normal to its level curve. A shared tangent therefore gives parallel gradients, possibly pointing in opposite directions.
25 · The multiplier equations give sides 5 and 5. Along the constraint, area is 25-(x-5)², proving the maximum is 25.
0 · Feasibility gives 0<=x<=7. The minimum of f=x is 0 at endpoint (0,7), which the interior multiplier equations do not find.
True · With this convention and a smooth regular optimal branch, V'(c)=lambda. The derivative may be negative.
8 · Substitution gives x²+y²=2(x-2)²+8. This is at least 8, with equality at x=y=2.
grad g is zero, so the regular theorem does not apply · At the origin, grad g=(0,0) but grad f=(1,0). No multiplier satisfies the equation. The required nonzero constraint gradient is absent.
3 · On the plane, the objective equals 3+(x-1)²+(y-1)²+(z-1)². The minimum is 3 at (1,1,1).
9 · The candidate is (1,2,2). On the plane, f=9+(x-1)²+(y-2)²+(z-2)², proving the minimum is 9.