Find the minimum of c(3a+4b) when a^2+b^2 +c^2=1.

[ssd]

Homework Statement

Given a,b,c real and a^2+b^2 +c^2=1, to find the minimum of c(3a+4b)

Homework Equations

The Attempt at a Solution

No positive clue yet.

 

[mastrofoffi]

What is the condition for a point $(a,b,c)$ to be the minimum point of a function?

 

[phyzguy]

I suggest you look up Lagrange's method of undetermined multipliers. It is a good method for solving problems of this type. Have you studied it in your class?

 

[Ray Vickson]

Homework Statement

Given a,b,c real and a^2+b^2 +c^2=1, to find the minimum of c(3a+4b)

Homework Equations

The Attempt at a Solution

No positive clue yet.

For each choice of $a$ and $b$ there are two choices for $c$:
$$c = + \sqrt{1-a^2-b^2}\\
c = - \sqrt{1-a^2-b^2}
$$
Each choice of $c = c(a,b)$ gives you a function $F(a,b) = (3a + 4b) c(a,b)$ to minimize. This new problem does not have an equality constraint anymore, but it still has a restriction $a^2 + b^2 \leq 1$, which you can try to ignore when making your first solution attempt

 

[ssd]

Thanks all for your kind suggestions. The Lagrangian multiplier solves the problem.

 

[mfb]

I would introduce new coordinates:
$$x=\frac {3a+4b}{5}$$
$$y=\frac {4a-3b}{5}$$
This is just a rotation in the a,b plane.
Now we have to minimize $5cx$ with the constraint $c^2+x^2+y^2=1$. As you found an answer already: By inspection, we see that $c=-x=\pm \frac{1}{\sqrt 2}$ leads to a minimum. Alternatively, introduce a second set of coordinates for the sum and the difference of c and x to make it a one-dimensional problem.

 

[ssd]

Thanks mfb. Nice approach.

 

[epenguin]

Thanks all for your kind suggestions. The Lagrangian multiplier solves the problem.

If that solves the problem what is the solution please?

I have some doubts, whether e.g. for the problem to make sense a, b, c should be restricted to be positive, or ask minimum absolute value or that your minimum is a a calculus minimum not the least value.

 

[ssd]

The result is a=-1/2, b=-1/2, c=1/√2 as per my calculations.

 

[phyzguy]

The result is a=-1/2, b=-1/2, c=1/√2 as per my calculations.

I think you made a mistake. I agree that c^2 = 1/2. The value of c(3a+4b) for your solution is -7√2/4 = -2.47. There is another condition where c(3a+4b) = -2.5. Why don't you walk us through your LaGrange multiplier calculation and let's see where you went wrong.

 

[ssd]

Surely I shall check my calculations. Notwithstanding though, when it is -2.5?

 

[mfb]

At $c=-x=\pm \frac{1}{\sqrt 2}$ and $y=0$ using x and y as defined in post 6. a and b will have different magnitudes at the minima.

 

[phyzguy]

Surely I shall check my calculations. Notwithstanding though, when it is -2.5?

Since this is in the homework section, I can't give you my solution. Like I asked, walk me through how you arrived at your solution and we will critique it.

 

[ssd]

Well, I got it. a=-3/5√2 and b= -4/5√2.
Actually I mentally calculated the minimum of 3a+4√(1/2-a^2) is at a=1/2 and that was the mistake. Since, our minimum could be the min of 3a-4√(1/2-a^2).