Science:Math Exam Resources/Courses/MATH110/April 2014/Question 04/Solution 1

We proceed using the standard template for solving optimisation questions. The exact steps might vary between course and section, but the overall steps will be similar.

1. A diagram.

2. So let's define the rectangular pasture to have length $l$ and width $w$. We want to minimise cost $C$.
3. Total cost of fencing will be $C=10w+10w+10l+15l=20w+25l$. We also need to have the area be $500=lw$.
4. Combining the two equations we get: $C(l)=25l+20\left({\frac {500}{l}}\right)=25l+{\frac {10000}{l}}$. In particular, the domain of the function $C$ is $l\in (0,\infty )$.
5. Differentiate, we get: $C'(l)=25-{\frac {10000}{l^{2}}}$. In the domain, the critical point is the solution of $25l^{2}=10000$ which is $l=\pm 20$. We disregard the negative solution not in the domain. So that means $l=20$. To see that it's a global min,

   Interval	$C'(l)$	INC/DEC
   $(0,20)$	$C'(10)<0$	DEC
   $(20,\infty )$	$C'(100)>0$	INC

   
   

   So that means $l=20$ is a local min. Since it is decreasing to the left, $C(20)$ must be lower than all other values to the left. By the same argument, it must be lower than all the points to the right as well. So this is indeed a global min.
6. The lowest cost for the fence can be reached by making the length of $20m$ and width $25m$.