Lagrange multipliers

**terulic** · January 4th 2010, 01:21 AM

I am having difficulty with the following problem :

A lecturer eats only two types of food and has a salary I > 0 . The quality of his lectures is given by x1^a1 * x2^a2 , where xi is the daily consumption of food i, respectively, and a i > 0 are constants, i = 1, 2 , a with a1 + a2 = 1 . The price of food i is pi > 0 , i = 1, 2 . The lecturer wishes to give lectures that are
of quality as high as possible, without spending more than his income. What is the optimal food consumption? If the salary is increased, after recent presidential elections, with ∆I , how much of an increase of the quality of lectures can be expected?

Can someone help please?

**HallsofIvy** · January 4th 2010, 04:28 AM

Originally Posted by terulic

I am having difficulty with the following problem :

A lecturer eats only two types of food and has a salary I > 0 . The quality of his lectures is given by x1^a1 * x2^a2 , where xi is the daily consumption of food i, respectively, and a i > 0 are constants, i = 1, 2 , a with a1 + a2 = 1 . The price of food i is pi > 0 , i = 1, 2 . The lecturer wishes to give lectures that are
of quality as high as possible, without spending more than his income. What is the optimal food consumption? If the salary is increased, after recent presidential elections, with ∆I , how much of an increase of the quality of lectures can be expected?

Can someone help please?

So your object function is $F= x_1^{a_1}x_2^{1-a_1}$ and you want to maximize that subject to $G= p_1x_1+ p_1x_2= I$ .
Now, $\nabla F= a_1x_1^{a_1-1}\vec{i}+ (1-a_1)x_2^{-a_1}\vec{j}$ and $\nabla G= p_1\vec{i}+ p_2\vec{j}$ .

Since you titled this "Lagrange multipliers", I assume you know that the optimum value will be where $\nabla F= \lambda\nabla G$ for some number $\lambda$ . In this case, that means $a_1x_1^{a_1-1}\vec{i}+ (1-a_1)x_2^{-a_1}\vec{j}= \lambda p_1\vec{i}+ \lambda p_2\vec{j}$ so you have the two equations $a_1x_1= \lambda p_1$ and $a_2x_2= \lambda p_2$ . I recommend dividing one equation by another to eliminate $\lambda$ , that you don't really care about, and then using the constraint $p_1x_1+ p_2x_2= I$ to solve for $x_1$ and $x_2$ . Of course, you answers will be in terms of $a_1$ , $a_2$ , $p_1$ , $p_2$ , and I.

To answer the second question, subtract the result with "I" from the result with " $I+ \Delta I$ ".

**terulic** · January 4th 2010, 04:30 AM

Thq, I will come back later today with the rest of the solution. I just needed a jump start.

**terulic** · January 14th 2010, 12:43 AM

Originally Posted by HallsofIvy

So your object function is $F= x_1^{a_1}x_2^{1-a_1}$ and you want to maximize that subject to $G= p_1x_1+ p_1x_2= I$ .
Now, $\nabla F= a_1x_1^{a_1-1}\vec{i}+ (1-a_1)x_2^{-a_1}\vec{j}$ and $\nabla G= p_1\vec{i}+ p_2\vec{j}$ .

Since you titled this "Lagrange multipliers", I assume you know that the optimum value will be where $\nabla F= \lambda\nabla G$ for some number $\lambda$ . In this case, that means $a_1x_1^{a_1-1}\vec{i}+ (1-a_1)x_2^{-a_1}\vec{j}= \lambda p_1\vec{i}+ \lambda p_2\vec{j}$ so you have the two equations $a_1x_1= \lambda p_1$ and $a_2x_2= \lambda p_2$ . I recommend dividing one equation by another to eliminate $\lambda$ , that you don't really care about, and then using the constraint $p_1x_1+ p_2x_2= I$ to solve for $x_1$ and $x_2$ . Of course, you answers will be in terms of $a_1$ , $a_2$ , $p_1$ , $p_2$ , and I.

To answer the second question, subtract the result with "I" from the result with " $I+ \Delta I$ ".

For the first question i got $x_1 = a_2p_1I/(p_1^{2}a_2+a_1p_2^{2})$ and similar for $x_2$ Do I need to substitute the values in $F= x_1^{a_1}x_2^{1-a_1}$ and then substract from the same with " $I+ \Delta I$ " ? What am i missing?

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