The mass fraction of, then y=k/ab, where k >；; 0 is the ratio

Example coefficient, according to the meaning of the question, that is, the values of a and b are the values of y.

Minimum. According to the topic, there are

4 b+ 2ab+2a = 60(a & gt； 0, b & gt0), ...4 points.

Get b = 30-a/2+a (0

So y=k/ab=k/((30a-a2)/(2+a))

=k/(-a+32-64/(a+2))

=k/(34-(a+2+64/(a+2))

≥k/(34-2)=k/ 18，

When a+2=64/(a+2), take the equal sign,

Y reaches the minimum value. ..... 8 points

In this case, a = 6, and a =- 10 (discarded). Substitute a=6 into ①.

Get b=3. Therefore, when a is 6 meters and b is 3 meters, after precipitation,

The mass fraction of this impurity in the water flowing out from the back is the smallest.

..... 12 point

Solution 2: according to the meaning of the question, that is, find the values of a and b to maximize ab.

Know from the title

4a+2ab+2a = 60(a & gt； 0, b & gt0), ...4 points.

That is, a+2b a+2 b+ab = 30(a >；; 0，b & gt0)。 ∫a+2b≥2，

∴2+ab≤30, if and only if a=2b, upward.

Type the equal sign. From a>0 and b>0, the solution is 0.

That is, when a=2b, ab takes the maximum value,

Its maximum value is 18. ..... 10 point

∴2b2= 18。 The solution is b=3 and a=6. So when a is 6 meters,

When b is 3m, the mass of impurities in the water flowing out after precipitation.

The quantity score is the smallest. ..... 12 point