Respuesta :
Area of the smallest region that can be photographed
[tex]\begin{gathered} =5\times7 \\ =35\operatorname{km}^2 \end{gathered}[/tex]Area which is 5 times its original size
[tex]\begin{gathered} =5\times35 \\ =175\operatorname{km}^2 \end{gathered}[/tex]If the length and width increases at a rate of 2km/sec, its area at any time t (where t is in seconds) will be:
[tex]Area=(5+2t)(7+2t)[/tex]Now, the time it takes for the area A to be at least 5 times its original size will be the time t when:
[tex](5+2t)(7+2t)=175[/tex]Next, we solve the equation derived above for t
[tex]\begin{gathered} \implies35+10t+14t+4t^2=175 \\ \implies4t^2+24t+35-175=0 \\ \implies4t^2+24t-140=0 \\ \implies4(t^2+6t-35)=0 \\ \text{Divide both sides by 4} \\ \implies t^2+6t-35=0 \end{gathered}[/tex]We can then solve for t using the quadratic formula
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex][tex]\begin{gathered} t=\frac{-6\pm\sqrt{6^2-4\times1\times-35}}{2\times1} \\ =\frac{-6\pm\sqrt{36+140}}{2} \\ =\frac{-6\pm\sqrt{176}}{2} \\ =\frac{-6\pm13.27}{2} \\ \text{Therefore:} \\ t=\frac{-6+13.27}{2}\text{ or }\frac{-6-13.27}{2} \\ t=3.64\text{ seconds or -9.64 seconds} \end{gathered}[/tex]Since time cannot be a negative value, the value t=-9.64 seconds is invalid.
Therefore: t=3.64 seconds
Therefore, it would take at least 3.64 seconds for the area, A to be at least 5 times its original size.
