Asked by Anonymous
(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?
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?
Answers
Answered by
Damon
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
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
Answered by
Anonymous
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