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shivamamity

  • 3 years ago

i have problem in calculating MRS (marginal rate of substitution) from MU(marginal utility). please explain with example ?

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  1. adnane.benali
    • 3 years ago
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    MRS of two goods x and y, say coke and orange juice, written as MRS of yx = (marginal utility from the change in x) divided by (marginal utility from the change in y). \[MRS _{yx} = \frac{ MU _{x} }{ MU _{y} }\] If you're given the MU of each good, then just plug them in. Say the utility function is \[U = x ^{2}y\] \[MU _{x} = \frac{ dU }{ dx } = 2xy \] \[MU _{y} = \frac{ dU }{ dy } = x ^{2}\] so \[MRS _{yx} = \frac{ MU _{x} }{ MU _{y} } = \frac{ 2xy }{ x ^{2} } = \frac{ 2y }{ x }\] Not sure if this helps you, but it's how I do it algebraically.

  2. shivamamity
    • 3 years ago
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    thanks for your support .....but i still have a problem in it. MUx=dU/dx=2xy 1-from where this 2xy came ? MUy=dU/dy=x power 2 2-from where this x power 2 came ? 3- Here dU/dx is change in marginal utility due to change in x. but here we don't know what was earlier MU of x and what is later MR utility of x due to change in x . So how it is done ? 4-we don't know earlier amount of x and later amount of x IF i don't know all this then how can it be calculated ? please help and explain each value from where it comes ? and same prob. in calculating MU of Y

  3. adnane.benali
    • 3 years ago
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    Ok. MUx=dU/dx is a concept in Algebra (did you do Algebra by any chance by the way?), it's called the "derivative" of the function U(x). The derivative in general tells us how much is U changing when x changes. In other words, how fast the curve of a function is going up or down. For example, function f(x)=x when plotted, is a line from the origin (0,0) going up with an angle of 45 degrees from the x axis, correct? this means that when x changes by 1, f(x) changes by 1 as well, and this is true for any two points along the plot. The change will never change. It is constant. It is 1. If we plot this change, with x on the x-axis, and delta-f(x) on the y-axis, we should get a straight line crossing the y-axis at 1 and parallel to the x-axis. This is called the derivative of the original function f(x)=x. The delta-f(x) is then written df(x)/dx or df/dx. So df/dx = 1 the way to determine a derivative in algebra is easy: 1. You take the exponent of x 2. Multiply it by the coefficient of x 3. decrement the exponent of x by 1 in general if f(x) = cx^3 then df(x)/dx = 3cx^2 in our example above we had f(x)=x so df(x)/dx = 1x^0 and x^0=1 so the whole thing = 1 Now, U = 2xy dU(x)/dx = d(2xy)/dx the only variable here is x. y is kept constant, called a parameter. So it is a constant just like 2. So, dU(x)/dx = d(2xy)/dx = 2yx^0 = 2y Similarly, dU(y)/dy = d(2xy)/dy = 2xy^0 = 2x terminology: dU(x)/dx = MUx and dU(y)/dy = MUy MRSyx = MUx/MUy = 2y/2x = y/x if U = x^3 y^5 MUx = dU(x)/dx = d(x^3 y^5)/dx = 3x^2 y^5 MUy = dU(y)/dy = d(x^3 y^5)/dy = 15x^3 y^4 so MRSyx = MYx/MUy = (3x^2 y^5)/(15x^3 y^4) = y/5x Hope this helps.

  4. shivamamity
    • 3 years ago
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    WOW this is great ......... actually i am not having maths back ground in my past .but from past 15 days i started studying these topic function ,limit, continuity . i will cover it in few month . but thanks again you helped me .from past 2 month i am studying chapter of consumer behavior and all the concept which i faced solved my this problem did nit able to solve I was very upset . but finally god send someone to help me .......

  5. adnane.benali
    • 3 years ago
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    Correction in the last example: MUy = dU(y)/dy = d(x^3 y^5)/dy = 5x^3 y^4 ---- not 15... so MRSyx = MYx/MUy = (3x^2 y^5)/(5x^3 y^4) = 3y/5x You're welcome! keep it up.

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