Rice production requires both labor and capital investments in equipment and land. Suppose that if x dollars per acre aer invested in labor and y dollars per acre are investd in equpiment and land, then the yield P of rice per acre is given by the formula P = 100 times the square root of x + a50 times the square root of y. If a farmer invests $40/acre, how hould he divide the $40 between labor and capital investment in order to maximize the amount produced?
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To determine how the farmer should divide the $40 investment between labor and capital investment in order to maximize the amount of rice produced, we need to analyze the given formula for the yield of rice per acre.
The formula for the yield P of rice per acre is given by:
P = 100√x + 50√y
Now, let's assign the amount the farmer invests in labor as L dollars, and the amount invested in equipment and land as C dollars. Since the total investment is $40, we have the equation:
L + C = $40
To maximize the yield of rice per acre, we need to find the values of L and C that maximize the formula P = 100√L + 50√C, given the constraint L + C = $40.
To solve this problem, we can use a mathematical technique called optimization. In this case, we will use the method of substitution.
First, express one of the variables in terms of the other using the constraint equation L + C = $40. Solve for C:
C = $40 - L
Next, substitute C in the formula for P:
P = 100√L + 50√($40 - L)
Now, let's graph this equation to visually observe the relationship between the investment in labor (L) and the yield of rice per acre (P). This will help us identify the maximum point on the graph.
Now, when we graph P = 100√L + 50√($40 - L), we will get a curve. The maximum point on this curve represents the maximum yield of rice per acre.
However, since we are dealing with a constrained optimization problem (L + C = $40), we can simplify the equation further by substituting C, as we did previously:
P = 100√L + 50√($40 - L)
P = 100√L + 50√($40 - L)
Simplifying further, we have:
P = 100√L + 50√($40 - L)
Next, we will differentiate P with respect to L to find the critical points. Taking the derivative of P with respect to L:
dP/dL = 100/(2√L) - 50/(2√($40 - L))
To find the critical points, set dP/dL = 0 and solve for L:
100/(2√L) - 50/(2√($40 - L)) = 0
Multiply through by 2√L√($40 - L):
100√($40 - L) - 50√L = 0
Divide through by 50:
2√($40 - L) - √L = 0
Split it to two terms:
2√($40 - L) = √L
Square both sides:
4($40 - L) = L
Expand and rearrange:
160 - 4L = L
Combine like terms:
5L = 160
Divide by 5:
L = 32
Now, substitute L = 32 into the constraint equation to find the investment in capital:
C = $40 - L
C = $40 - 32
C = $8
Therefore, the farmer should invest $32 on labor and $8 on capital in order to maximize the amount of rice produced, given the $40 total investment.