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Step 1: Formulate the corresponding maximum problem using the budget restriction.
Let $p_1 $ and $p_2 $ be the prices for respectively the products 1 and 2, and let $ y $ be the available budget (suppose you spend this amount $y$ completely on these products). Then the problem can be stated as
\\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} && p_1 \\cdot q_1 + p_2 \\cdot q_2 = 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} && p_1 \\cdot q_1 + p_2\\cdot q_2 = 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 ( y - p_1 \\cdot q_1 - p_2\\cdot q_2 )
\\end{eqnarray}
Step 3: Determine the first order conditions in order to calulate 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 p_1 = 0 \\\\
\\mbox{(2) } \\quad \\frac{\\partial {L}}{\\partial q_2} & = & {\\frac{\\var{n1min1}}{\\var{n1}}} \\cdot {\\frac {1} {q_2-{\\var{a2}}}} - \\lambda \\cdot p_2 = 0 \\\\
\\mbox{(3) } \\quad \\frac{\\partial {L}}{\\partial \\lambda} & = & y - p_1 \\cdot q_1 - p_2 \\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} {p_1 \\cdot(q_1-{\\var{a1}})} }& = & {\\frac{\\var{n1min1}}{\\var{n1}}}\\cdot{\\frac {1} {p_2 \\cdot(q_2-{\\var{a2}})}}\\\\ \\\\
{p_2} \\cdot {(q_2-{\\var{a2}})} & = & {\\var{n1min1}} \\cdot {p_1} \\cdot {(q_1-{\\var{a1}})} \\\\ \\\\
\\mbox{(4)} \\quad \\quad \\quad {p_2} \\cdot {q_2} & = & {\\var{a2}} \\cdot {p_2} + {\\var{n1min1}} \\cdot {p_1} \\cdot {(q_1-{\\var{a1}})}
\\end{eqnarray}
i.e.
\\begin{eqnarray}
{q_2} & = & {\\var{a2}} + {\\frac {1} {p_2}} \\cdot {\\var{n1min1}} \\cdot {p_1} \\cdot {(q_1-{\\var{a1}})}
\\end{eqnarray}
Substitute equation (4) in the budget restriction:
\\begin{eqnarray}
{p_1} \\cdot {q_1} + {\\var{a2}} \\cdot {p_2} + {\\var{n1min1}} \\cdot {p_1}\\cdot {(q_1-{\\var{a1}})} & = & y\\\\
{p_1} \\cdot {q_1} + {\\var{a2}} \\cdot {p_2} + {\\var{n1min1}} \\cdot {p_1} \\cdot{q_1} - {\\var{n1min1}} \\cdot {\\var{a1}} \\cdot {p_1} & = & y\\\\
{\\var{n1}} \\cdot {p_1}\\cdot {q_1} & = & y + {\\var{n1min1maala1}} \\cdot {p_1} - {\\var{a2}} \\cdot {p_2} \\\\
{q_1} & = & {\\frac {1}{\\var{n1} \\cdot {p_1}}} \\cdot { ( y + {\\var{n1min1maala1}} \\cdot {p_1} - {\\var{a2}} \\cdot {p_2} ) } \\\\
{q_1} & = & {\\frac {y + {\\var{n1min1maala1}} \\cdot {p_1} - {\\var{a2}} \\cdot {p_2}}{\\var{n1} \\cdot {p_1}}}
\\end{eqnarray}
Substituting this last formula for $q_1$ in the equation (4) leads to a formula for $q_2$:
\\begin{eqnarray}
{q_2} & = & \\frac{\\var{t2}\\cdot y-\\var{a1maalt2} \\cdot {p_1} + \\var{a2}\\cdot{p_2}}{\\var{n1} \\cdot {p_2}}
\\end{eqnarray}
Answer.
The demand functions are
\\begin{eqnarray}
{q_1 \\left( p_1,p_2,y\\right)} & = & {\\frac {y + {\\var{n1min1maala1}} \\cdot {p_1} - {\\var{a2}} \\cdot {p_2}}{\\var{n1} \\cdot {p_1}}}\\\\
{q_2\\left( p_1,p_2,y\\right)} & = & \\frac{\\var{t2}\\cdot y-\\var{a1maalt2}\\cdot{p_1} + \\var{a2}\\cdot {p_2}}{\\var{n1} \\cdot {p_2}}
\\end{eqnarray}
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