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A Mixture Problem A tank initially contains 10 gal of brine with 2 lb of salt. Brine with 1.5 lb of salt per gallon enters the tank at the rate of 3 gal/min, and the well-stirred mixture leaves the tank at the rate of 4 gal/min. It can be shown that the amount of salt in the tank after $t$ min is $x$ lb, where $$x=f(t)=1.5(10-t)-0.0013(10-t)^{4} \quad(0 \leq t \leq 10)$$ What is the maximum amount of salt present in the tank at any time?

          A Mixture Problem A tank initially contains 10 gal of brine with 2 lb of salt. Brine with 1.5 lb of salt per gallon enters the tank at the rate of 3 gal/min, and the well-stirred mixture leaves the tank at the rate of 4 gal/min. It can be shown that the amount of salt in the tank after $t$ min is $x$ lb, where
$$x=f(t)=1.5(10-t)-0.0013(10-t)^{4} \quad(0 \leq t \leq 10)$$
What is the maximum amount of salt present in the tank at any time?

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Added by Elizabeth P.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Linda Hand

07/15/2022

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5 - 0.0052(10-t)^3 \] ** Show more…

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A Mixture Problem A tank initially contains 10 gal of brine with 2 lb of salt. Brine with 1.5 lb of salt per gallon enters the tank at the rate of 3 gal/min, and the well-stirred mixture leaves the tank at the rate of 4 gal/min. It can be shown that the amount of salt in the tank after $t$ min is $x$ lb, where $$x=f(t)=1.5(10-t)-0.0013(10-t)^{4} \quad(0 \leq t \leq 10)$$ What is the maximum amount of salt present in the tank at any time?

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Transcript

00:01 All right, so this equation x gives us the amount of salt in the tank at any time t.

00:08 And this part right here just says this is the equation for t between zero and 10 minutes or hours or seconds or whatever it was.

00:18 Then after the 10, we don't know what happens.

00:22 Okay, now the question is find the maximum amount of x in there.

00:26 So to find the maximum of anything, take the derivative, set up.

00:31 It equal to zero and find the critical points and then check to see which one gives you the maximum.

00:37 All right.

00:38 So x prime or dxt, oops, x prime.

00:44 So it's 1 .5 times minus 1 because that's the derivative of 10 minus t, minus 0 .0013 times 4 times 10 minus t to the third times minus 1.

00:59 So minus 1 .5 plus 0 .0052, 10 minus t to the third equals 0 .0052.

01:19 10 minus t to the third equals 1 .52.

01:20 10 minus t to the third equals 1 .5.

01:24 10 minus t to the third equals 1 .5 over 0 .0052.

01:34 10 minus t to the third.

01:36 Equals the cube root of 1 .5 over 0 .052.

01:44 And then we'll subtract 10 and then divide or multiply by minus 1.

01:49 So t will be 10 minus the cube root of this.

01:57 So let me see if i can calculate that.

02:05 Okay, 1 .5 divided by 0 .005 to cube root minus minus that plus 10, 3 .39262.

02:32 Okay, so that's the critical value, 0, 3 .39262.

02:40 10.

02:42 All right, so now what we have to do is make sure that's the maximum...

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