At the start, the tank contains
(0.25 lb/gal) * (100 gal) = 25 lb
of sugar. Let [tex]S(t)[/tex] be the amount of sugar in the tank at time [tex]t[/tex]. Then [tex]S(0)=25[/tex].
Sugar is added to the tank at a rate of P lb/min, and removed at a rate of
[tex]\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}[/tex]
and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,
[tex]\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}[/tex]
Separate variables, integrate, and solve for S.
[tex]\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt[/tex]
[tex]\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt[/tex]
[tex]-100\ln\left|P-\dfrac S{100}\right|=t+C[/tex]
[tex]\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t[/tex]
[tex]P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}[/tex]
[tex]\dfrac S{100}=P-Ce^{-100t}[/tex]
[tex]S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}[/tex]
Use the initial value to solve for C :
[tex]S(0)=25\implies 25=100P-C\implies C=100P-25[/tex]
[tex]\implies S(t)=100P-(100P-25)e^{-100t}[/tex]
The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time
[tex]1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}[/tex]
has passed. At this time, we want the tank to contain
(0.5 lb/gal) * (5 gal) = 2.5 lb
of sugar, so we pick P such that
[tex]S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}[/tex]
