#11. There are $320 available to fence in rectangular garden. The fencing for the side of the garden facing the road costs $6 per foot and the fencing for the other three sides costs $2 per foot. Consider the problem of finding the dimensions of the largest possible garden.
So I solved everything and got:
A=xy as the objective equation
C=8y+4x=320 as the constraint equation
A=(80x-x^2)/2
A'=(80-2x)/2 then I set that to zero and got x=40, y=20.
How come the correct answer is x=20 y=40? More importantly, how do you get the domain? (which is 0<x<40)
* i realized that I set C=8y+4x=320 while the answer key has C=8x+4y=320* Would my answer be still correct?
If you write 8y+4x=320 then you will need to plug in y=40. In this case, we have the restriction 0 < y < 40. All you did was switch the variable names.
The letter "x" isn't special, it's just a convention to name an independent variable x and a dependent variable y.
So we can validly write x = f(y) = (320-8y)/4 and ask for the domain (keeping in mind there is a requirement that x > 0), then find that the domain is 0 < y < 40.
Make sense?
Okay, you haven't responded yet, but I'm worried that maybe you're still confused, so maybe this will help.
We can look at this problem a few different ways. For completeness, I'll just type out all 4 ways so hopefully you see how it all fits.
Situation 1: We chose to write 8x+4y=320, and we let y = f(x).
This is the book answer.
In this case, the domain is 0 < x < 40 and the range is 0 < y < 80.
Situation 2: We chose to write 8x+4y=320, and we let x = f(y).
Then the domain is 0 < y < 80 and the range is 0 < x < 40.
Situation 3: We chose to write 8y+4x=320, and we let y = f(x).
Then the domain is 0 < x < 80 and the range is 0 < y < 40.
Situation 4: We chose to write 8y+4x=320, and we let x = f(y).
Then the domain is 0 < y < 40 and the range is 0 < x < 80.
See how it's always the same, but with certain things switched with each other.
I hope it's clear now. If not, just ask.