Brewery Filtration Window with a Parameterised Rate Model
A small brewery models its filtration rate by R(t)=(t^2+at+8)/(t+b) litres per hour, where t is hours after start, 0≤t≤12, and a,b are positive constants. During commissioning, the rate is 6 L/h at t=2 and 7 L/h at t=5. The operations manager wants the filtration rate kept at or below 9 L/h to avoid clogging the plate filter. A colleague solves R(t)=9 by multiplying through by t+b, obtains two positive roots, and proposes to accept both without checking the interval 0≤t≤12 or the sign of R(t)-9. You are asked to review the model and the proposed operating window. The brewery can adjust start time only within the 12-hour batch; any interval must be expressed in hours after start. The manager also wonders whether the denominator can vanish for an admissible t and whether that changes the algebraic solution.
1. Determine a and b from the two commissioning measurements, and write R(t) in quotient form. State the denominator condition on 0≤t≤12.
Read the worked answer
Substitute t=2 and R=6: (4+2a+8)/(2+b)=6, so 12+2a=12+6b, hence a=3b. Substitute t=5 and R=7: (25+5a+8)/(5+b)=7, so 33+5a=35+7b, hence 5a-7b=2. Using a=3b gives 15b-7b=2, so b=1/4 and a=3/4. The model is R(t)=(t^2+(3/4)t+8)/(t+1/4). Since b=1/4>0, the denominator t+1/4 is positive for every t in [0,12]; there is no excluded point in the operating interval. Long division gives R(t)=t+1/2+63/[8(t+1/4)], which is useful for sign checks because the last term is positive on the whole interval.
2. Solve R(t)≤9 on 0≤t≤12. Show the sign chart and explain why the two positive roots do not automatically give the acceptable interval.
Read the worked answer
Set R(t)=9: (t^2+(3/4)t+8)/(t+1/4)=9. Because t+1/4>0 on the interval, multiplying through is reversible: t^2+(3/4)t+8=9t+9/4, so 4t^2-33t+23=0. The roots are t=[33±sqrt(1089-368)]/8=[33±sqrt721]/8. Numerically these are about 0.768 and 7.481. The quadratic 4t^2-33t+23 opens upward, so R(t)-9 is positive outside the roots and negative between them. Thus the acceptable window is [33-sqrt721]/8 ≤ t ≤ [33+sqrt721]/8. The endpoint checks prevent a common error: at t=0, R(0)=32>9, and at t=12, R(12)>9. Therefore the two positive roots are the boundaries of one interval inside [0,12], not separate acceptable intervals. The denominator never vanishes, so no extra restriction comes from it.