Respuesta :
SOLUTION
Step1: write out the giving equation
[tex]\frac{dp}{dt}=50t^2-100t^{\frac{3}{2}}[/tex]Step2: Integrate both sides of the equation above
[tex]\int \frac{dp}{dt}=\int 50t^2dt-\int 100t^{\frac{3}{2}}dt[/tex]Then simplify by integrating both sides
[tex]p(t)=\frac{50t^{2+1}}{2+1}-\frac{100t^{\frac{3}{2}+1}}{\frac{3}{2}+1}+c[/tex][tex]p(t)=\frac{50}{3}t^3-40t^{\frac{5}{2}}+c[/tex]since the initial value is 25,000, then
the Population function is
[tex]\begin{gathered} p(t)=\frac{50}{3}t^3-40t^{\frac{5}{2}}+25000\ldots\ldots..\ldots\text{.. is the population function} \\ \text{where t=time in years} \end{gathered}[/tex]b). For the population to reach 50,000 the time will be
[tex]\begin{gathered} 50000=\frac{50}{3}t^3-40t^{\frac{5}{2}}+2500 \\ 50000-25000=\frac{50}{3}t^3-40t^{\frac{5}{2}} \\ 25000=\frac{50}{3}t^3-40t^{\frac{5}{2}} \\ \text{Then} \\ \frac{50}{3}t^3-40t^{\frac{5}{2}}-25000=0 \\ \end{gathered}[/tex]Multiply the equation by 3, we have
[tex]\begin{gathered} 50t^3-120t^{\frac{5}{2}}-75000=0 \\ \end{gathered}[/tex]To solve this we rewrite the function as
[tex]14400t^5=\mleft(-50t^3+75000\mright)^2[/tex]The value of t becomes
[tex]\begin{gathered} t\approx\: 15.628,\: t\approx\: 9.443 \\ t=15.625\text{ satisfy the equation above } \end{gathered}[/tex]Then it will take approximately
[tex]16\text{years}[/tex]
