Prunella raises peaches where L is the number of units of labor she uses and T is the number of units of land she uses

Prunella raises peaches where L is the number of units of labor she uses and T is the number of units of land she uses. Her output is f(L,T) = LT bushels of peaches. If the cost of land is 17 and the wage for labor is 2, what would be the slope of an iso-cost line, where total cost is $500. T is on the vertical axis.

The Correct Answer and Explanation is:

To solve this, we need to understand the concept of an iso-cost line and the equation for it. An iso-cost line shows all combinations of labor (L) and land (T) that result in the same total cost. The formula for total cost (C) is:C=wL+pTC = wL + pTC=wL+pT

Where:

  • www is the wage rate for labor (2),
  • ppp is the price of land (17),
  • LLL is the amount of labor used,
  • TTT is the amount of land used.

The total cost is given as $500, so we set up the equation:500=2L+17T500 = 2L + 17T500=2L+17T

Now, we need to solve for LLL in terms of TTT, which will give us the equation of the iso-cost line:2L=500−17T2L = 500 – 17T2L=500−17TL=500−17T2L = \frac{500 – 17T}{2}L=2500−17T​

Now, we can find the slope of the iso-cost line. The slope of the iso-cost line is given by the rate at which the cost of land and labor trade off. Mathematically, it is the ratio of the price of labor to the price of land:slope=−wp\text{slope} = -\frac{w}{p}slope=−pw​

Substituting the values for www and ppp:slope=−217\text{slope} = -\frac{2}{17}slope=−172​

Thus, the slope of the iso-cost line is −217-\frac{2}{17}−172​, or approximately −0.1176-0.1176−0.1176. This means that for every additional unit of land, Prunella must use 0.1176 fewer units of labor to maintain the same total cost of $500.

In summary, the slope of the iso-cost line represents the trade-off between labor and land while keeping costs constant. In this case, the negative slope indicates that increasing the amount of land used requires reducing the amount of labor to keep the total cost fixed at $500.

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