Suppose \(U=(x, y)=4 x^{2}+3 y^{2}\) a. \(\quad\) Calculate \(\partial U / \partial x, \partial U / \partial y\) b. Evaluate these partial derivatives at \(x=1, y=2\) c. Write the total differential for \(U\) d. Calculate \(d y / d x\) for \(d U=0\) - that is, what is the implied trade-off between \(x\) and \(y\) holding \(U\) constant? e. Show \(U=16\) when \(x=1, y=2\) f. In what ratio must \(x\) and \(y\) change to hold \(U\) constant at 16 for movements away from \(x=1, y=2 ?\) g. More generally, what is the shape of the \(U=16\) contour line for this function? What is the slope of that line?

Short Answer

Expert verified
In summary, we found the partial derivatives of U with respect to x and y, evaluated them at the given point (x=1, y=2), and determined the total differential for U. Additionally, we found the implied trade-off between x and y to hold U constant, verified that U=16 for the given point, and calculated the shape of the U=16 contour line and its slope. The results show that at the point (x=1, y=2), the partial derivatives are 8 and 12 for x and y, respectively. The total differential is \(8x dx + 6y dy\), and the slope of the contour line at (x=1, y=2) is -2/3.

Step by step solution

01

Calculate \(\frac{\partial U}{\partial x}\)

To find the partial derivative of U with respect to x, treat y as a constant and differentiate U with respect to x: \(\frac{\partial U}{\partial x} = 8x\).
02

Calculate \(\frac{\partial U}{\partial y}\)

To find the partial derivative of U with respect to y, treat x as a constant and differentiate U with respect to y: \(\frac{\partial U}{\partial y} = 6y\). b. Evaluating Partial Derivatives
03

Evaluate \(\frac{\partial U}{\partial x}\) at \((x=1, y=2)\)

Evaluate the partial derivative with respect to x at the given point: \(\left.\frac{\partial U}{\partial x}\right\vert_{(x=1,y=2)} = 8(1) = 8\).
04

Evaluate \(\frac{\partial U}{\partial y}\) at \((x=1, y=2)\)

Evaluate the partial derivative with respect to y at the given point: \(\left.\frac{\partial U}{\partial y}\right\vert_{(x=1,y=2)} = 6(2) = 12\). c. Total Differential
05

Write the total differential for U

The total differential for U is given by: \(dU = \frac{\partial U}{\partial x} dx + \frac{\partial U}{\partial y} dy = 8x dx + 6y dy\). d. Implied Trade-off
06

Calculate \(\frac{dy}{dx}\) for \(dU=0\)

Set the total differential equal to zero, and solve for \(\frac{dy}{dx}\): \(0 = 8x dx + 6y dy \Rightarrow -8x dx = 6y dy \Rightarrow \frac{dy}{dx} = -\frac{4x}{3y}\). e. Show U=16
07

Verify that U=16 when \((x=1, y=2)\)

Substitute x=1 and y=2 into the function U: \(U=4(1)^2+3(2)^2=4+12=16\). f. Ratio of x and y
08

Find the ratio of x and y to hold U constant

Use the previously found formula for \(\frac{dy}{dx}\) to find the ratio for holding U constant: \(\frac{dy}{dx} = -\frac{4x}{3y}\). g. Contour Line and Slope
09

Find the shape of the U=16 contour line

The contour line for U=16 is the set of all points (x y) that satisfy the equation \(4x^2 + 3y^2 = 16\). This is an ellipse.
10

Find the slope of the U=16 contour line

Use the formula for \(\frac{dy}{dx}\) at the given point(x=1, y=2) to find the slope: \(\frac{dy}{dx} = -\frac{4(1)}{3(2)} = -\frac{2}{3}\). The slope of the contour line at the point (x=1, y=2) is -2/3.

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