Maximize a Function: Extrema of ρsinθ + cosθ over [0,pi/2]

[Qube]

Homework Statement

Find the extrema of over [0,pi/2]

ρsinθ + cosθ = f(θ), where ρ is a constant

Homework Equations

Extrema occur at critical points and global extrema occur at end points.

The Attempt at a Solution

f(0) = 1
f(pi/2) = ρ

f'(θ) = ρcosθ - sinθ = 0

ρcosθ = sinθ.

θ = pi/4. This seems to maximize the original function because geometrically, the largest triangle is constructed on the unit circle by making the two legs (the cosine and sine functions) equal in length.

f(pi/4) = √2ρ > ρ.

Or is θ = arctan(ρ)?

 

[ehild]

Homework Statement

Find the extrema of over [0,pi/2]

ρsinθ + cosθ = f(θ), where ρ is a constant

Homework Equations

Extrema occur at critical points and global extrema occur at end points.

The Attempt at a Solution

f(0) = 1
f(pi/2) = ρ

f'(θ) = ρcosθ - sinθ = 0

ρcosθ = sinθ.

[STRIKE]θ = pi/4. This seems to maximize the original function because geometrically, the largest triangle is constructed on the unit circle by making the two legs (the cosine and sine functions) equal in length.
f(pi/4) = √2ρ > ρ.
[/STRIKE]

No, the derivative is not zero at θ=pi/4. Not the area of a triangle in the unit circle has to be maximized but the f(θ) function. And d(pi/4) is not √2ρ.

Or is θ = arctan(ρ)?

yes, it is. Is there a local maximum or a minimum? What is f(θ) at θ=arctan(ρ)?

ehild

 

[Ray Vickson]

Homework Statement

Find the extrema of over [0,pi/2]

ρsinθ + cosθ = f(θ), where ρ is a constant

Homework Equations

Extrema occur at critical points and global extrema occur at end points.

The Attempt at a Solution

f(0) = 1
f(pi/2) = ρ

f'(θ) = ρcosθ - sinθ = 0

ρcosθ = sinθ.

θ = pi/4. This seems to maximize the original function because geometrically, the largest triangle is constructed on the unit circle by making the two legs (the cosine and sine functions) equal in length.

f(pi/4) = √2ρ > ρ.

Or is θ = arctan(ρ)?

I don't know if your statement "Extrema occur at critical points and global extrema occur at end points." is intended to apply to ANY function, or just to the one given in the question. In any case that statement is incorrect in general; depending on the sign of ρ, the statement can be true in some cases and false in other cases. You do not even need calculus to solve this problem; just use the standard trick of applying trigonometric addition formulas, to write f as
f(t) = R \sin(t+w), \text{ where }\: R = \sqrt{1+\rho^2}, \; \cos(w) = \frac{\rho}{R}, \: \sin(w) = \frac{1}{R}.
What would the graph of this f(t) look like on the interval [0,π/2]?