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What practical lessons can be drawn from the factor factor relationship in the combination of soybean meal and fishmeal for the formulation of layer mash?

Economics

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Here's the solution for question 1c.iii and 1c.iv:

The production function is given as y=1.5x+0.0084x2−0.000046x3y = 1.5x + 0.0084x^2 - 0.000046x^3, where yy is the Total Physical Product (TPP) and xx is the input (maize).

From the previous derivations (or by calculating them now):
The Average Physical Product (APP) function is:

APP=yx=1.5x+0.0084x2−0.000046x3x=1.5+0.0084x−0.000046x2\text{APP} = \frac{y}{x} = \frac{1.5x + 0.0084x^2 - 0.000046x^3}{x} = 1.5 + 0.0084x - 0.000046x^2

The Marginal Physical Product (MPP) function is:

MPP=dydx=ddx(1.5x+0.0084x2−0.000046x3)=1.5+2(0.0084)x−3(0.000046)x2\text{MPP} = \frac{dy}{dx} = \frac{d}{dx} (1.5x + 0.0084x^2 - 0.000046x^3) = 1.5 + 2(0.0084)x - 3(0.000046)x^2 MPP=1.5+0.0168x−0.000138x2\text{MPP} = 1.5 + 0.0168x - 0.000138x^2

1c.iii) Construct a table for the three functions TPP, APP and MPP for the values of x (maize) ranging from 0 – 480 at an interval of 40 units.

  1. Calculate TPP, APP, and MPP for each value of xx.

    We will calculate the values for x=0,40,80,120,160,200,240,280,320,360,400,440,480x = 0, 40, 80, 120, 160, 200, 240, 280, 320, 360, 400, 440, 480.

    x (Maize Units)TPP (Eggs)APP (Eggs/Unit)MPP (Eggs/Unit)
    00.001.501.50
    4070.501.761.95
    80150.211.881.96
    120221.401.851.53
    160266.621.670.66
    200268.001.34-0.66
    240207.940.87-2.42
    28068.770.25-4.62
    320-167.17-0.52-7.26
    360-517.54-1.44-10.35
    400-1000.00-2.50-13.86
    440-1632.22-3.71-17.82
    480-2431.87-5.07-22.24

    1c.iv) With this production function, how would you advice a poultry farmer to operate within a particular range of the variable input x (maize).

  2. Identify the stages of production.

    • Stage I (Increasing Returns): TPP increases at an increasing rate, MPP is increasing, and APP is increasing. This stage ends where APP is maximized (and APP = MPP).
    • Stage II (Diminishing Returns): TPP increases at a decreasing rate, MPP is positive but decreasing, and APP is decreasing. This stage ends where TPP is maximized (and MPP = 0). This is the rational stage of production.
    • Stage III (Negative Returns): TPP decreases, and MPP is negative. It is irrational to operate in this stage.
  3. Determine the boundaries of the stages.

    • End of Stage I (APP is maximized): Set APP = MPP. 1.5+0.0084x−0.000046x2=1.5+0.0168x−0.000138x21.5 + 0.0084x - 0.000046x^2 = 1.5 + 0.0168x - 0.000138x^2 0.0084x−0.000046x2=0.0168x−0.000138x20.0084x - 0.000046x^2 = 0.0168x - 0.000138x^2 0=0.0084x−0.000092x20 = 0.0084x - 0.000092x^2 0=x(0.0084−0.000092x)0 = x(0.0084 - 0.000092x) This gives x=0x=0 or x=0.00840.000092≈91.3x = \frac{0.0084}{0.000092} \approx 91.3 units.
    • End of Stage II (TPP is maximized, MPP = 0): Set MPP = 0. 1.5+0.0168x−0.000138x2=01.5 + 0.0168x - 0.000138x^2 = 0 Using the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}: x=−0.0168±(0.0168)2−4(−0.000138)(1.5)2(−0.000138)x = \frac{-0.0168 \pm \sqrt{(0.0168)^2 - 4(-0.000138)(1.5)}}{2(-0.000138)} x=−0.0168±0.00028224+0.000828−0.000276x = \frac{-0.0168 \pm \sqrt{0.00028224 + 0.000828}}{-0.000276} x=−0.0168±0.00111024−0.000276x = \frac{-0.0168 \pm \sqrt{0.00111024}}{-0.000276} x=−0.0168±0.0333202−0.000276x = \frac{-0.0168 \pm 0.0333202}{-0.000276} The relevant positive root is x=−0.0168−0.0333202−0.000276≈181.59x = \frac{-0.0168 - 0.0333202}{-0.000276} \approx 181.59 units.
  4. Provide advice to the farmer.

    The farmer should operate in Stage II of production. This stage occurs when the input of maize (xx) is between approximately 91.3 units and 181.59 units. In this range, the total output of eggs (TPP) is increasing, and the average output per unit of maize (APP) is still positive, although it is declining. Operating beyond 181.59 units would lead to a decrease in total egg production, making it economically irrational.

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