About Cody Thompson Expertise I can answer most questions in Calc I, Calc II, Calc III, Algebra, and Trigonometry. Also, don't hesitate to send me a Probability/Stats problem.
I have knowledge in Diff. Eq. problems(separation of variables, Newton's cooling law, emptying a tank, etc), but can't make a guarantee.
My math knowledge is 'jack of all trades, master of none' type deal. I know a little number theory, linear algebra also, but as before, you can ask but I can't make any promises.
Experience I currently teach Intermediate Algebra and Statistics at a community college in Western Md.
I was a field engineer/surveyor for years, therefore, I have used mathematics in 'real-world' situations.
Organizations Mathematical Association of America.
Education/Credentials Bachelor's degree in Mathematics.
Question A man wants to build a rectangular enclosure for his herd. He only has $900 to spend on the fence and wants the largest size for his money. He plans to build the pen along the river on his property, so he does not have to put a fence on that side. The side of the fence parallel to the fence will cost $5 per foot to build, whereas the sides perpendicular to the river will cost $3 per foot.What dimensions should he choose?
Answer Hello Chris:
The perimeter of the fence is l+2w. The parallel side has cost $5 per foot so we have 5l. The perp sides to the river have cost $3 per foot, so we have 3*2w.
The cost is then 5l+6w=900...........[1]
Area is l*w...........[2]
Solve [1] for, say, l:
l=(900-6w)/5
Sub into [2] and simplify:
180w-((6w^2)/5)
Differentiate and get: 180-(12w/5)
Set to 0 and solve for w and get w=75
Plug this back into the equation above and we find l=90
