Question

Max    u=q₁^6 q₂^4+1.5Ln(q₁ )+Ln(q₂)

s.to

3q₁+4q₂=100

Answer 1

According to the question it is necessary to maximize the function $u=q_{1}^6q_{2}^4+1.5ln(q_{1})+ln(q_{2})$ taking into account the following restriction equation $3q_{1}+4q_{2}=100$

ANSWER

The first step is the Langrangian function (L) creation.

This function is the following:

$L(q_{1},q_{2},\lambda)=u(q_{1},q_{2})+\lambda*(3q_{1}+4q_{2}-100)$

The second step is to take the partial derivatives of the Lagrangian with respect to q₁ and q_2. After that using the restriction equation and the equations that the partial Langrangian derivatives are equal to zero we can find the stationary points of L.

$dL/dq_{1}=6*q_{1}^5q_{2}^4+1.5/q_{1}+3\lambda=0$ (1)

$dL/dq_{2}=4q_{1}^6q_{2}^3+1/q_{2}+4\lambda=0$ (2)

We can multyply the equation (2) by a q₂:

$4q_{1}^6q_{2}^4+1+4\lambda*q_{2}=0$ (2.1)

We can multyply the equation (1) by a q₁:

$6*q_{1}^6q_{2}^4+1.5+3\lambda*q_{1}=0$ (1.1)

The equation (1.1) can be rewritten as follows:

$q_{1}^6q_{2}^4=-1.5/6-3\lambda*q_{1}/6=-0.25-\lambda*q_{1}/2$ (1.2)

Taking into account the (1.2) and (2.1) we can write:

$-1-2\lambda*q_{1}+1+4\lambda*q_{2}=0$

and hence $q_{1}=2q_{2}$ (1.3)

Using the restriction equation we can find q₁ and q₂

$3*2q_{2}+4q_{2}=100$

Hence, $q_{2}=10$ and $q_{1}=20$. It is stationary point (that does not depend on $\lambda$ ).

The third step

To prove that the values q₂=10 and q₁=20 are solution we need to prove that $d_{2}L$ is negative in the point q₂=10 and q₁=20

$d^2L=d^2L/dq_{1}^2*dq_{1}^2+2d^2L/(dq_{1}dq_{2})*dq_{1}dq_{2}+d^2L/dq_{2}^2*dq_{2}^2$ (3.1)

$d^2L/dq_{1}^2=30q_{1}^4q_{2}^4-1.5/q_{1}^2=a$ (3.2)

$d^2L/dq_{2}^2=12q_{1}^6q_{2}^2-1/q_{2}^2=b$ (3.3)

$2d^2L/(dq_{1}dq_{2})=24q_{1}^5q_{2}^3=c$ (3.4)

Taking into account (3.2), (3.3) and (3.4) the equation (3.1) can be rewritten as follows:

$d^2L=a*dq_{1}^2+2b*dq_{1}dq_{2}+c*dq_{2}^2$ (3.1.1)

Based on the restriction equation we can write that

$3dq_{1}=-4dq_{2}$

Hence. (3.1.1) can be rewritten as follows

$d^2L=a*dq_{1}^2-2*3/4b*dq_{1}dq_{1}+c*dq_{1}^2*9/16$ (3.1.2)

Then,

$d^2L=(a-1.5b+c*9/16)dq_{1}^2$ (3.1.2).

Let's calculate the a, b, and c while q₂=10 and q₁=20.

When q₂=10 and q₁=20, $a=30*100*160,000-1.5/400=480,000,000-1.5/400$

When q₂=10 and q₁=20, $b=12*64,000,000*100-0.01=76,800,000,000-0.01$

When q₂=10 and q₁=20, $c=24*1,000*3,200,000= 76,800,000,000$

Based on the values of a, b, and c we can conclude that while q₂=10 and q₁=20 the $d_{2}L$ is less than zero. Hence, the point q₂=10 and q₁=20 is point where the value of the utilitu function (denoted by u) is the highest.