I need help with Qs 4, 5 and 6!!
- Three sides of a rectangular paddock are to be fenced, the fourth side being an existing straight water drain. If 1000m of fencing is available, what dimensions should be used for the paddock so that it encloses the maximum area possible?
I have filled up around 4 pages of working out for these questions. I believe if someone could talk me through q 4 then I will understand q 5 and 6
Thanks in advance
Call the side perpendicular to the drain $x$, and the other side $y$. The area will then be $A=xy$. Because of the constraint that you only have 1000m of fence but only need to make three sides of the paddock, the perimeter of the paddock not including the water drain will be 1000m, so that $2x+y=1000$. Rearranging for $y$ we get $y=1000-2x$ and the area becomes $$A(x)=x(1000-2x) = 1000x-2x^2$$ We could do this using calculus: for a turning point, ${dA\over dx}=0$ and so $${dA\over dx}=1000-4x = 0$$ giving $x=250m$. Since it's a quadratic though we can determine the maximum value by completing the square on $A(x)$: $$A(x)=1000x-2x^2=-2(x^2-500x)=-2[(x-250)^2-250^2]$$ which has a maximum value when $x=250$, and so $A_{max}=2\times250^2=125000.$