I'm gonna have abunch of questions, cause I have a test. >_<
If you planened to fence a rectangular pasture that must contain 4000 sq meters, what dimensions would require the least amount of fencing?
Perimeter= 2l + 2w
area= lw=4000
or l=4000/w
P= 2(4000/w + w)
Now minimize this function, solve for w.
To find the dimensions that require the least amount of fencing for a rectangular pasture, you need to minimize the function you derived, P = 2(4000/w + w).
Let's go through the steps to solve for the value of w that minimizes this function:
1. Start with the function: P = 2(4000/w + w)
2. Simplify the function: P = 8000/w + 2w
3. Take the derivative of P with respect to w to find the critical points. The derivative of the function is dP/dw = -8000/w^2 + 2.
4. Set the derivative equal to zero and solve for w: -8000/w^2 + 2 = 0
5. Multiply both sides of the equation by w^2 to eliminate the fraction: -8000 + 2w^2 = 0
6. Rearrange the equation: 2w^2 = 8000
7. Divide both sides by 2: w^2 = 4000
8. Take the square root of both sides: w = sqrt(4000)
9. Simplify the square root: w = 20√10
Now that you have the value of w, you can substitute it back into the equation for P to find the corresponding value of the perimeter, which will give you the dimensions that require the least amount of fencing.