an orchard involving a particular type of fruit tree the average weight of fruit w kg produced per tree depends on N the number of trees per 100 m square according to the rule w=(600-15n) for 11<n<26 using calculus and clearly justifying that your value would indeed give a maximum determine the value of n that gives the maximum total weight of this fruit that produced per 100 m square
total fruit per square is just #trees * avg weight per tree, or
f(n) = n*w(n) = 600n-15n^2
f'(n) = 600 - 30n
maximum occurs when f'(n)=0, or n = 20
Note that n=20 lies in the domain of the function.
You can either use what you know about parabolas to realize that this one opens down, so any f'=0 will be a maximum.
Or, you can use f"(n)=-30 to show that the curve is concave down, making the extremum a max.
To determine the value of n that gives the maximum total weight of fruit produced per 100 square meters in the orchard, we can use calculus to find the maximum of the function w = (600 - 15n).
Step 1: Find the derivative of the function with respect to n.
dw/dn = -15
Step 2: Set the derivative equal to zero and solve for n to find the critical point(s).
-15 = 0
There is no value of n that satisfies this equation since -15 is a constant.
Step 3: Examine the endpoints of the given range, 11 < n < 26.
The given range is 11 < n < 26. Since the function w = (600 - 15n) is defined for this range, we need to evaluate the function at the endpoints to check if they yield maximum values.
w(11) = 600 - 15(11) = 600 - 165 = 435 kg
w(26) = 600 - 15(26) = 600 - 390 = 210 kg
Step 4: Compare the values of w at the endpoints to determine the maximum.
From the calculations, we can see that w(11) > w(26). Therefore, the maximum total weight of fruit produced per 100 square meters occurs when n = 11 in the given range.
To find the value of N that gives the maximum total weight of fruit produced per 100 m square, we need to maximize the function that relates the average weight of the fruit per tree, w, to the number of trees, N.
The given rule is:
w = 600 - 15N
First, we need to determine the total weight of fruit produced per 100 m square. Let's call this function T(N). It can be calculated by multiplying the average weight of fruit per tree by the number of trees per 100 m square.
T(N) = w * (100/N)
Substituting the equation for w from the given rule, we get:
T(N) = (600 - 15N) * (100/N)
To find the value of N that maximizes T(N), we need to find the critical point of this function. The critical point occurs when the derivative of T(N) with respect to N is equal to zero.
So, let's find the derivative of T(N) with respect to N:
T'(N) = d/dN [(600 - 15N) * (100/N)]
T'(N) = (d/dN [600 - 15N]) * (100/N) + (600 - 15N) * (d/dN [100/N])
Differentiating each term separately:
= (-15) * (100/N) + (600 - 15N) * (-100/N^2)
Simplifying further:
= -1500/N + (600N - 15N^2) * (-100/N^2)
Multiplying through by -1/N^2:
= -1500/N - 600N/N^2 + 15N^2/N^2
Combining like terms:
= (-1500 - 600N + 15N^2) / N^2
Now we set T'(N) equal to zero and solve for N:
(-1500 - 600N + 15N^2) / N^2 = 0
Multiplying through by N^2:
-1500 - 600N + 15N^2 = 0
Now we can solve this quadratic equation for N. We can use the quadratic formula:
N = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, a = 15, b = -600, and c = -1500.
N = (600 ± sqrt((-600)^2 - 4 * 15 * (-1500))) / (2 * 15)
Simplifying further:
N = (600 ± sqrt(360000 + 90000)) / 30
N = (600 ± sqrt(450000)) / 30
N = (600 ± 300√5) / 30
N = 20 ± 10√5
Since n represents the number of trees, and the original equation w = (600 - 15n) is defined for 11 < n < 26, we need to consider the values of N within this range. The value N = 20 + 10√5 is greater than 26, so we discard it.
Therefore, the value N = 20 - 10√5 gives the maximum total weight of fruit produced per 100 m square.