Since both x and y have to be positive numbers, you simply have to find for what values 200 = 4x+3y lies in the first quadrant.
Find the x and y intercepts
let x = 0 , y = 200/3
let y = 0 , x = 50
so 0 < x < 50
0 < y < 200/3
Hello, please help me.
There is a question where I have to optimize the area of a field that is being fence in like this: [|] i.e. two rectangle fields, side by side: area= 2x by y. I get to use 200ft of fencing. From that diagram and the amount of fencing I have, I made this equation: 200 = 4x +3y and A=2xy
from that area equation, I solved for A'(x) and got the critical number x=25 and noted that it is a (relative?) maximum.
From that, I solved for A subbing in x=25 and y=100/3 into A=2xy, where I got 1666.67 ft^2
Now this is where I stuck. I want to apply the absolute extrema test, but I'm not sure how to find out what the closed interval is. I'm pretty sure it has to do with the equation 200=4x +3y, but I have no idea how to go about it. Could you please show me how?
Again, thank you everyone who has helped me study for this test, I really appreciate it.
1 answer
1 answer