Transcription
Let's work on question six in this video. So we spent 500 euros on two goods: food and clothing. I guess the utility is given to us: F² + C². So that matters. The price of food is five, and the price of clothing is ten. How much should we buy? Okay, so let's set up our Lagrangian and play a bit with the math. It's pretty, pretty mechanical, but it's wise to get really used to it.
So we maximize the function, the utility function F² + C² - λ times our constraint. And let's write the constraint here so we know what that is. The price of food is five times the units of food plus the price of clothing is ten times the units of clothing equals the income of 500. So we put that here: 5F + 10C - 500. And now we take the partial derivatives with respect to food and then with respect to clothing, make that equals to zero.
So the partial derivative of the Lagrangian with respect to F: F² derivative with respect to F is just 2F. Here we have nothing with F, and then -λ x 5. The derivative we keep the constants only, so λ times five. And then 10C and 500 are constants; their derivatives are just zero. So that'll be it. Now we do the same with respect to clothing. The Lagrangian derivative with respect to clothing: F² is a constant because F and anything else is a constant; it's just zero. C to the power two derivative becomes just 2C - λ times this one goes away because it's a constant, and then 10C derivative is just 10. So 10 = 0.
Now again, we can see we have λ on both sides, so that should ring a bell that we will do something about it. So let's transfer them on each side: 2F = λ times 5, and here we have 2C = λ times 10. Let's leave λ on one side only. So let's actually go below so we have more space to write. We have now 2F / 5 = λ from the first, and then 2C / 10 = λ from the second. We can do some simplification here: two and ten they cancel out, so two and then we have only five. Meaning that now you have λ = 2/5 times F, and here λ = C/5, or we could write it as 1/5 times clothing. Now λ = λ, which means that 2/5 times F = 1/5 times clothing. And again we can cancel out something: five and five go away because they're both denominators on both sides, and it's an equation, so we can do it: 2 times F = 2 times clothing.
And now this relationship we will substitute it or plug it in our budget constraint, and our budget constraint is over here. So we take this and put it over there. Now let's move to the right to have space and solve that: 5 times F, so we got 5 times F + 10 times C = 500. Instead of clothing, we'll use 2 times F. So we have 5 times F + 10 times 2 times F, because that's the same as the clothing, equals to 500. So now we have 5F + 10 times 2 = 20F = 500. 5 + 20 = 25F = 500. And the optimal consumption of food is 500 divided by 25, which is equal to 20 units. So that's 20 units; that's one answer. And how can we calculate the clothing? Well, just by substituting because clothing is twice as much. So clothing will be equal to two times 20, which is equal to 40. And that's it; we are done.