Step 1: Formulate the corresponding maximum problem using the budget restriction.
\\begin{eqnarray}
\\max_{q_1,q_2} U(q_1,q_2) &=& {\\var{c1}} \\cdot (q_1-{\\var{a1}})^{1 / {\\var{n1}}} \\cdot (q_2-{\\var{a2}})^{{\\var{n1min1}} / {\\var{n1}}}
\\mbox{ if } && \\var{p1} \\cdot q_1 + \\var{p2} \\cdot q_2 = \\var{y}
\\end{eqnarray}
or equivalently
\\begin{eqnarray}
\\max_{q_1,q_2} ln U(q_1,q_2) &=& ln({\\var{c1}}) + {\\frac{1}{\\var{n1}}} \\cdot ln(q_1-{\\var{a1}}) + {\\frac{\\var{n1min1}}{\\var{n1}}} \\cdot ln(q_2-{\\var{a2}}) \\\\
\\mbox{ if } && \\var{p1} \\cdot q_1 + \\var{p2} \\cdot q_2 = \\var{y}
\\end{eqnarray}
Step 2: Write down the corresponding Lagrange function for this constrained maximum problem.
\\begin{eqnarray}
L(q_1, q_2, \\lambda) &=& ln U(q_1,q_2) + \\lambda \\cdot (y - p_1 \\cdot q_1 - p_2 \\cdot q_2) \\\\
&=& ln({\\var{c1}}) + {\\frac{1}{\\var{n1}}} \\cdot ln(q_1-{\\var{a1}}) + {\\frac{\\var{n1min1}}{\\var{n1}}} \\cdot ln(q_2-{\\var{a2}}) + \\lambda \\cdot ( \\var{y} - \\var{p1} \\cdot q_1 - \\var{p2}\\cdot q_2 )
\\end{eqnarray}
Step 3: Determine the first order conditions in order to calculate critical points of the Lagrangian.
\\begin{eqnarray}
\\mbox{(1) } \\quad \\frac{\\partial {L}}{\\partial q_1} & = & {\\frac{1}{\\var{n1}}} \\cdot {\\frac {1} {q_1-{\\var{a1}}} } - \\lambda \\cdot \\var{p1} = 0 \\\\
\\mbox{(2) } \\quad \\frac{\\partial {L}}{\\partial q_2} & = & {\\frac{\\var{n1min1}}{\\var{n1}}} \\cdot {\\frac {1} {q_2-{\\var{a2}}}} - \\lambda \\cdot \\var{p2} = 0 \\\\
\\mbox{(3) } \\quad \\frac{\\partial {L}}{\\partial \\lambda} & = & \\var{y} - \\var{p1} \\cdot q_1 - \\var{p2} \\cdot q_2 = 0
\\end{eqnarray}
Step 4: Find the solution of this system of equations in $q_1$, $q_2$ and $\\lambda$.
Calculating $\\lambda$ in both equations (1) and (2) leads to
\\begin{eqnarray}
{\\frac{1}{\\var{n1}}}\\cdot{\\frac {1} {\\var{p1}} \\cdot (q_1-{\\var{a1}}) }& = & {\\frac{\\var{n1min1}}{\\var{n1}}}\\cdot{\\frac {1} {\\var{p2} \\cdot(q_2-{\\var{a2}})}}\\\\
{\\var{p2}} \\cdot {(q_2-{\\var{a2}})} & = & {\\var{n1min1}} \\cdot {\\var{p1}} \\cdot {(q_1-{\\var{a1}})} \\\\
{\\var{p2}} \\cdot {q_2} & = & {\\var{a2}} \\cdot {\\var{p2}} + {\\var{n1min1}} \\cdot {\\var{p1}} \\cdot {(q_1-{\\var{a1}})} = {\\var{a2*p2}} + {\\var{n1min1*p1}} \\cdot {(q_1-{\\var{a1}})}
\\end{eqnarray}
i.e.
\\begin{eqnarray}
\\mbox{(4)} \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad{q_2} & = & {\\var{a2-n1min1*s*a1}} + {\\var{n1min1*s}} \\cdot {q_1}
\\end{eqnarray}
Substitute(4) in the budget restriction (3):
\\begin{eqnarray}
{\\var{p1}} \\cdot {q_1} + {\\var{(a2-n1min1*s*a1)*p2}} + {\\var{n1min1*s*p2}} \\cdot {q_1} & = & \\var{y}\\\\
{q_1} + {\\var{a2-n1min1*a1}} + {\\var{n1min1}} \\cdot {q_1} & = & \\var{n1*v+a1+a2}\\\\
{\\var{n1}} \\cdot {q_1} + {\\var{a2-n1min1*a1}} & = & \\var{n1*v+a1+a2}\\\\
q^*_1 &=& \\var{q1}
\\end{eqnarray}
Substituting this last formula in equation (4), we can calculate $q_2$:
\\begin{eqnarray}
{q^*_2} & = & \\var{q2}
\\end{eqnarray}
Answer.
For
\\begin{eqnarray}
q^*_1&=&\\var{q1} \\\\
q^*_2&=&\\var{q2}
\\end{eqnarray}
the maximal utility value is
\\[ U (q^*_1,q^*_2) \\approx \\var{nutfinaal} \\]
{toonfiguur()}
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