(a) If a farmer plants x units of wheat in his field, 0 ≤ x ≤ 144, the yield will be 12x − (x^2/12) units.
How much wheat should he plant for the maximum yield?
(b) In the problem above, it costs the farmer $108 for each unit of wheat he plants, and he is able to sell each unit he harvests for $72. How much should he plant to maximize his profit?
y = 12 x - x^2/12
12 y = 144 x - x^2
x^2 -144 x = -12 y
where is the vertex?
x^2 - 144 x + 5184 = -12 y + 5184
(x-72)^2 = -12 ( y - 32)
max yield of 32 at x = 72
p = -108 x + 72 (12 x -x^2/12)
p = -108 x + 864 x - 6 x^2
p = -6 x^2 + 756 x
where is the vertex?
6 x^2 - 756 x = -p
x^2 - 126 x = -p/6
x^2 - 126 x + 3969 = -p/6 + 3969
(x-63)^2 = -(1/6) (p - 23814)
max profit of $23,814 at 63 units planted
Thank you!
(a) To find the amount of wheat the farmer should plant for the maximum yield, we need to find the value of x that maximizes the function 12x - (x^2/12).
Step 1: Find the derivative of the function with respect to x.
f'(x) = 12 - (2/12)x
Step 2: Set the derivative equal to zero and solve for x to find the critical points.
12 - (2/12)x = 0
12 = (2/12)x
12 * 12 = 2x
144 = 2x
x = 72
Step 3: Evaluate the second derivative to determine if the critical point is a maximum or minimum.
f''(x) = -2/12 = -1/6
Since the second derivative is negative, the critical point x=72 is a maximum.
Therefore, the farmer should plant 72 units of wheat for the maximum yield.
(b) To find the amount the farmer should plant to maximize his profit, we need to consider the cost and selling price of the wheat.
Step 1: Calculate the profit function.
Profit(x) = Revenue - Cost
Profit(x) = 72x - 108x = -36x
Step 2: Find the derivative of the profit function.
Profit'(x) = -36
Step 3: Since the derivative is a constant, there are no critical points to consider. The profit function is linear, and the farmer can maximize his profit by maximizing the amount of wheat planted.
Therefore, the farmer should plant the maximum amount possible, which is 144 units of wheat, to maximize his profit.
To find the amount of wheat the farmer should plant for the maximum yield, we can use calculus to find the critical points.
(a) To find the critical points, we differentiate the yield function with respect to x and set it equal to zero.
Y(x) = 12x - (x^2/12)
Y'(x) = 12 - (2x/12)
Setting Y'(x) to zero and solving for x:
12 - (2x/12) = 0
12 = 2x/12
2x = 12*12
x = 12*12 / 2
x = 72
Therefore, the farmer should plant 72 units of wheat for the maximum yield.
(b) To find the amount of wheat the farmer should plant to maximize his profit, we need to consider the cost and revenue as well. Profit is given by the equation:
Profit(x) = Revenue(x) - Cost(x)
Revenue(x) = selling price * yield = $72 * (12x - (x^2/12))
Cost(x) = cost per unit * x = $108 * x
Profit(x) = $72 * (12x - (x^2/12)) - $108 * x
To find the critical points, we differentiate the profit function with respect to x and set it equal to zero.
Profit'(x) = 72 * (12 - (2x/12)) - 108
Set Profit'(x) equal to zero and solve for x:
72 * (12 - (2x/12)) - 108 = 0
72 * (12 - (2x/12)) = 108
12 - (2x/12) = 108 / 72
12 - (2x/12) = 1.5
2x / 12 = 12 - 1.5
2x / 12 = 10.5
2x = 10.5 * 12
x = (10.5 * 12) / 2
x = 63
Therefore, the farmer should plant 63 units of wheat to maximize his profit.