# Maximization Problems Calculus

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Help with Calculus 1 – Maximization problem.?

What is the thought process when looking at this problem? How do I set up the equations in order to solve?

A viticulturist estimates that if 50 grapevines are planted per acre, each vine will produce 140 pounds of grapes. Each additional grapevine planted per acre (up to 20 extra vines) reduces the average yield per vine by 2 pounds. The viticulturist should plant ______ grapevines to maximize the yield per acre.

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60 vines will maximize the yield.

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We’re looking at planting 50 to 70 grape vines per acre.
We want to consider the change in yield per acre with respect to a change in the number of vines planted per acre.

Let y = total yield per acre
Let x = number of extra vines (over 50) that can be planted.
Then, what we’re looking for is dy/dx: The change in yield per acre with respect to the change in number of vines to be planted per acre; which is dependent upon the number of extra vines x that can be planted.

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(The yield per acre) = (the number of vines)•(average yield per vine)
y = the yield per acre and
50 + x = the number of vines to be planted per acre

As for the average yield per vine:
With 50 vines, the average yield would be 140 pounds per vine. But, with each additional vine x, the average goes down by 2 pounds.
So that the average loss per vine is 2x pounds: ie. with 1 extra vine the average drops by 2•1=2pounds. With 2 extra vines the average drops by 2•2=4 pounds, and with 3 extra vines the average drops by 6 pounds, etc.
Therefore,
140-2x = average yield per vine
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Now, then, by substitution:
(yield per acre) = (the number of vines)•(average yield per vine)

y = (50 + x)(140 – 2x)
y = -2x² + 40x + 7000
This is a parabola that opens down, since the coefficient of x² is negative.
It will, therefore, have a maximum when dy/dx = 0
Now,
dy/dx = -4x + 40 = 0 for x=10

Therefore, the viticulturist should plant 50+10=60 vines per acre.

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3.7a2 Optimization Problems – Calculus

Category: Calculus Problems
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