Question 1192675
I am assuming it is y=(2.8x)/(0.4x^2+1.3).
x=5, y=14/(10+1.3)=14/11.3=1.24
x=10, y=28/41.3=0.68
x=20, y=56/161.3=0.35
x=30, y=84/361.3=0.23
x=60, y=168/1441.3=0.117
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y=0.4=(2.8x)/(0.4x^2+1.3)
0.16x^2+0.52=2.8x
0.16x^2-2.8x+0.52=0
16x^2-280x+52=0; 4x^2-70x+13=0
x=17.3 min
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y=0.2
0.08x^2+0.26=2.8x
0.08x^2-2.8x+0.26=0
8x^2-280x+26=0; 4x^2-140x+13=0
x=34.91 min
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y=0.1
0.04x^2+0.13=2.8x
0.04x^-2.8x+0.13=0
4x^2-280x+13=0
x=70 min.
An hour and a half would be too long. The minimum is reached in 1h10m
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y=0.01
0.004x^2-2.8x+0.013=0, 695 minutes

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Substitute 2 for y 

2=(2.8x)/(0.4x^2+1.3)
0.8x^2+2.6=2.8x
0.8x^2-2.8x+2.6=0
8x^2-28x+26=0
4x^2-7x+13=0, and this has complex roots, so it does not reach 2 mg/l

Take the derivative and it is y'=(0.4x^2+1.3)*2.8-(2.8x)(0.8x)/(0.4x^2+1.3)^2
set y'=0 and multiply through by the denominator, which disappears
so 1.12x^2+3.64=2.24x^2 and 1.12x^2=3.64 and x=1.80 min maximum

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The horizontal lines are y=0.1,0.2,0.4,0.6,0.8 mg/l
{{{graph(300,300,-10,100,-2,2,0.1,0.2,0.4,0.6,0.8,(2.8x)/(0.4x^2+1.3))}}}