The variables u and v take positive values suc that (1/u) + (1/v) = 1/20. If s = u + v, find the least value of s as u varies

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- May 8th 2005, 09:56 AMJingXionGCalculus
The variables u and v take positive values suc that (1/u) + (1/v) = 1/20. If s = u + v, find the least value of s as u varies

- May 8th 2005, 11:56 PMticbol
"...find the least value of S if u varies...."

Relating to functions, this can ve viewed as S = f(u), or S is a function of u. S is the dependent variable and u is the independent variable.

So, if dS/du = 0, then we can solve for maximum S or minimum S.

We need to find dS/du.

S = u +v

Differentiate both sides with respect to u,

dS/du = du/du +dv/du

dS/du = 1 +dv/du ....(1)

We need to find dv/du.

1/u +1/v = 1/20

Differentiate both sides with respect to u,

[u(0) -1(1)]/(u^2) +[v(0) -1(dv/du)]/(v^2) = 0

-1/(u^2) -(dv/du)/(v^2) = 0

-(dv/du)/(v^2) = 1/(u^2)

Multiply both sides by (-v^2),

dv/du = [1/(u^2)]*(-v^2)

dv/du = -(v^2)/(u^2) ....***

Substitute that into (1),

dS/du = 1 +[-(v^2)/(u^2)]

ds/du = 1 -[(v^2)/(u^2)]

Set that to zero,

0 = 1 -[(v^2)/(u^2)]

v^2 / u^2 = 1

v^2 = u^2

v = u

That means, when u = v, the dS/du is 0.

That means also then that S is either maximum or minimum when u = v.

There are two popular ways to find if S is maximum or minimum when dS/du = 0.

One way is by finding the concavity of the graph of S at the point where dS/du is zero. That is by finding if the 2nd derivative of S with respect to u, at that point where dS/du = 0, is negative or positive.

S is maximum if the second derivative is negative.

S is minimum if the second derivative is positive.

You can do this if you like, if it is taught to you in school already. You will find that S'', or the 2nd derivative of S, is positive, so S is minimum.

(I can show how to find the S'' if you like.)

Another way is to investigate the slopes right before and right after the point where ds/du = 0. That is to investigate the signs of the 1st derivative of S, or S', right before and right after the point where ds/du = 0. [At the point where dS/du = 0, we found u=v.] Like if v=5, see the signs of dS/du at points (u=4 while v=5) and (u=6 while v=5).

If the signs change from negative to positive, then S is minimum.

If the signs change from positive to negative, then S is maximum.

dS/du = 1 +dv/du ....(1)

ds/du = 1 -[(v^2)/(u^2)]

When u=4 while v=5,

dS/du = 1 -[5^2 / 4^2] = 1 -25/16 = -9/16

It is negative.

When u=6 while v=5,

dS/du = 1 -[5^2 / 6^2] = 1 -25/36 = 11/36

It is positive.

Negative to positive, therefore, S is minimum, or least.

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So what is S then when u=v?

1/u +1/v = 1/20

Change v to u,

1/u +1/u = 1/20

2/u = 1/20

Cross multiply,

u = 2*20 = 40

And so, v = u = 40 also.

Therefore, the least S if u varies is

S = u +v

least S = 40 +40 = 80. ...answer. - May 9th 2005, 04:19 AMJingXionGThanks alot
Appreciate it :)