11. Find the average value of the function f(x) = 5 + 6e^(-3x) on the interval [

Question

11. Find the average value of the function f(x) = 5 + 6e^(-3x) on the interval [0,4]. 12. A farmer wishes to fence off a rectangular plot of land, using an existing wall as one of the sides. The total area enclosed must be 900 square yards. The fence on the side parallel to the wall will cost $30 per yard; the fences on the other two sides will cost $50 per yard. (a) What should the dimensions be in order to minimize the total cost of the fence? First draw and label an appropriate diagram. (b) Explain why your answer to (a) will give a minimum.

Explanation / Answer

f(x) = 5 + 6e^(-3x) over [0 , 4]

1/(4 - 0) * integral from 0 to 4 f(x)

1/4 * integral 0 to 4 [5 + 6e^(-3x)]

1/4 * [5x - 2e^(-3x)] from 0 to 4

1/4 * [(20 - 2e^-12) - (0 - 2)]

1/4 * [20 - 2/e^12 + 2]

1/4 * [22 - 2/e^12]

11/2 - 1/(2e^12)

12)

Length = x
Width = y

xy = 900 --> x = 900/y

Cost equation : x*30 + 2y*50 = 30x + 100y --> to be minimized

C = 30(900/y) + 100y

dC/dy = -27000/y^2 + 100 = 0

27000/y^2 = 100

y^2 = 270

y = 3sqrt(30)

x = 900/(3sqrt30)

x = 300/sqrt30

x = 10sqrt(30)

So, the side parallel to the wall must be of 10sqrt(30) yards
And the other two parallel sides must be of 3sqrt(30) yards in length

b)

d^2C/dy^2 = 5400/y^3 + 100

When y = 3sqrt(30) is plugged in, we get d^2C/dy^2 = positive value

So, we know that those values obtained in part A lead to the minimum

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