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Chapter 7: Problem 35

Compute \(\frac{\partial^{2} f}{\partial x^{2}},\) where \(f(x, y)=60 x^{3 / 4} y^{1 / 4},\) a production function (where \(x\) is units of labor). Explain why \(\frac{\partial^{2} f}{\partial x^{2}}\) is always negative.

Short Answer

Expert verified
The second partial derivative of \(f\) with respect to \(x\) is \[-\frac{45}{4} x^{-5/4} y^{1/4}\]. It is always negative because the derivative includes a negative constant multiplier.

Step by step solution

01

Determine the first partial derivative of f with respect to x

Given the function \[f(x, y) = 60 x^{3/4} y^{1/4}\], compute the first partial derivative with respect to x. Use the power rule for differentiation.\[\frac{\partial f}{\partial x} = 60 \cdot \frac{3}{4} x^{-1/4} y^{1/4} = 45 x^{-1/4} y^{1/4}\]
02

Determine the second partial derivative of f with respect to x

Now, compute the second partial derivative by differentiating the first partial derivative with respect to x again.\[\frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) = \frac{\partial}{\partial x} \left( 45 x^{-1/4} y^{1/4} \right)\]Apply the power rule for differentiation:\[\frac{\partial^{2} f}{\partial x^{2}} = 45 \cdot (-1/4) x^{-5/4} y^{1/4} = -\frac{45}{4} x^{-5/4} y^{1/4}\]
03

Explain why \(\frac{\partial^{2} f}{\partial x^{2}}\) is always negative

Observe the second derivative calculated in step 2: \[\frac{\partial^{2} f}{\partial x^{2}} = -\frac{45}{4} x^{-5/4} y^{1/4}\].Given that \(x > 0\) and \(y > 0\), both \(x^{-5/4}\) and \(y^{1/4}\) are positive. The constant multiplier, \(-\frac{45}{4}\), is negative. Therefore, the entire expression is always negative regardless of the values of \(x\) and \(y\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Differentiation
Partial differentiation is a fundamental concept in calculus where we focus on how a multivariable function changes with respect to one variable at a time, while keeping the other variables constant.
For example, if you have a function like \( f(x, y) \), you can compute the partial derivatives with respect to \( x \) and \( y \).
This gives us insight into the rate of change along each variable's direction.
The key formula for partial differentiation with respect to \( x \) for our function \( f(x, y) = 60 x^{3/4} y^{1/4} \):
\(\frac{\frac{d}{\frac{\frac{\frac{dx}}{0}}}}\frac{\frac{\frac_fx\frac{\frac{\frac{dx}\frac}}}}{dx}}}{\)\ By using the power rule of differentiation, we find that:
\( \frac{\frac{45 x^{-1/4}}{45 x^{-1/4}} x^{-1/4} y^{1/4}}}{45 x^{-1/4}\frac}\frac}{\) \ This gives us the rate of change of \( f \) with respect to \( x \).
Remember, we still have \( y \) as a constant during this process.
Production Function
A production function represents the relationship between the inputs a business uses and the output it creates.
In our case, the function \( f(x, y) = 60 x^{3/4} y^{1/4} \) describes how labor (\( x \)) and another input (\( y \) like capital) combine to produce output.
This particular form showcases diminishing returns to scale, where increasing one input results in smaller increases in output.
Production functions help businesses understand how changes in labor or capital affect their output.
It鈥檚 essential in economics and business management for optimizing resource allocation.
Negative Second Derivative
The second partial derivative tells us about the concavity of the function. In our exercise, we found the second partial derivative of the production function with respect to \( x \):
\( \frac{\frac{-459}{4} y^{1/4}.\frac}{\)\ This is always negative because:
  • \

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Most popular questions from this chapter

The value of residential property for tax purposes is usually much lower than its actual market value. If \(v\) is the market value, the assessed value for real estate taxes might be only \(40 \%\) of \(v .\) Suppose that the property \(\operatorname{tax}, T,\) in a community is given by the function $$T=f(r, v, x)=\frac{r}{100}(.40 v-x)$$ where \(v\) is the market value of a property (in dollars), \(x\) is a homeowner's exemption (a number of dollars depending on the type of property), and \(r\) is the tax rate (stated in dollars per hundred dollars). (a) Determine the real estate tax on a property valued at \(\$ 200,000\) with a homeowner's exemption of \(\$ 5000,\) assuming a tax rate of \(\$ 2.50\) per hundred dollars of net assessed value. (b) Determine the tax duc if the tax rate increases by \(20 \%\) to \(\$ 3.00\) per hundred dollars of net assessed value. Assume the same property value and homeowner's exemption. Does the tax due also increase by \(20 \% ?\)

Table 6 gives the U.S. minimum wage in dollars for certain years. $$\begin{array}{ll} \text {Table 6 U.S. Federal Minimum Wage} \\ \hline \text { Year } & \text { Wage } \\ \hline 2000 & \$ 5.15 \\ 2005 & \$ 5.15 \\ 2010 & \$ 7.25 \\ 2016 & \$ 7.25 \\ \hline \end{array}$$ (a) Use the method of least squares to obtain the straight line that best fits these data. [Hint: First convert Year to Years after 2000 .] (b) Estimate the minimum wage for the year 2008 . (c) If the trend determined by the straight line in part (a) continues, when will the minimum wage reach \(\$ 10 ?\)

The function \(f(x, y)=\frac{1}{2} x^{2}+2 x y+3 y^{2}-x+2 y\) has a minimum at some point \((x, y) .\) Find the values of \(x\) and \(y\) where this minimum occurs.

Find the values of \(x, y\) that minimize \(x^{2}+x y+y^{2}-2 x-5 y\) subject to the constraint \(1-x+y=0.\)

Both first partial derivatives of the function \(f(x, y)\) are zero at the given points. Use the second-derivative test to determine the nature of \(f(x, y)\) at each of these points. If the second derivative test is inconclusive, so state. $$f(x, y)=y e^{x}-3 x-y+5 ;(0,3)$$
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