Answer
(a) We show that:
$\nabla g\left( {x\left( c \right),y\left( c \right)} \right)\cdot\left( {x'\left( c \right),y'\left( c \right)} \right) = 1$
(b) We show that:
$\frac{d}{{dc}}f\left( {x\left( c \right),y\left( c \right)} \right) = \lambda $
(c) by interpreting $g\left( {x\left( c \right),y\left( c \right)} \right) = c$ as a "budget" constraint, we conclude that $\lambda$ is the rate of increase in $f$ per unit increase in the "budget level" $c$.
Work Step by Step
(a) We are given the constraint $g\left( {x\left( c \right),y\left( c \right)} \right) = c$.
We differentiate the equation $g\left( {x\left( c \right),y\left( c \right)} \right) = c$ with respect to $c$ using the Chain Rule (Theorem 1 of Section 15.6):
$\frac{{\partial g}}{{\partial c}} = \frac{{\partial g}}{{\partial x}}\frac{{\partial x}}{{\partial c}} + \frac{{\partial g}}{{\partial y}}\frac{{\partial y}}{{\partial c}} = 1$
Using dot product, we can write:
$\frac{{\partial g}}{{\partial x}}\frac{{\partial x}}{{\partial c}} + \frac{{\partial g}}{{\partial y}}\frac{{\partial y}}{{\partial c}} = \left( {\frac{{\partial g}}{{\partial x}},\frac{{\partial g}}{{\partial y}}} \right)\cdot\left( {\frac{{\partial x}}{{\partial c}},\frac{{\partial y}}{{\partial c}}} \right)$
So,
$\frac{{\partial g}}{{\partial c}} = \left( {\frac{{\partial g}}{{\partial x}},\frac{{\partial g}}{{\partial y}}} \right)\cdot\left( {\frac{{\partial x}}{{\partial c}},\frac{{\partial y}}{{\partial c}}} \right) = 1$
Since $\nabla g = \left( {\frac{{\partial g}}{{\partial x}},\frac{{\partial g}}{{\partial y}}} \right)$, $x'\left( c \right) = \frac{{\partial x}}{{\partial c}}$ and $y'\left( c \right) = \frac{{\partial y}}{{\partial c}}$, so
$\frac{{\partial g}}{{\partial c}} = \left( {\nabla g} \right)\cdot\left( {x',y'} \right) = 1$
Hence, $\nabla g\left( {x\left( c \right),y\left( c \right)} \right)\cdot\left( {x'\left( c \right),y'\left( c \right)} \right) = 1$.
(b) At the point $P$, we may write the function $f$ as $f\left( {x\left( c \right),y\left( c \right)} \right)$.
We differentiate the function $f\left( {x\left( c \right),y\left( c \right)} \right)$ with respect to $c$ using the Chain Rule:
$\frac{{df}}{{dc}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial c}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial c}}$
Similar to part (a) we get
$\frac{{df}}{{dc}} = \left( {\nabla f} \right)\cdot\left( {x',y'} \right)$
Writing in full we have
$\frac{d}{{dc}}f\left( {x\left( c \right),y\left( c \right)} \right) = \nabla f\left( {x\left( c \right),y\left( c \right)} \right)\cdot\left( {x'\left( c \right),y'\left( c \right)} \right)$
Using the Lagrange condition $\nabla {f_P} = \lambda \nabla {g_P}$, we get
$\frac{d}{{dc}}f\left( {x\left( c \right),y\left( c \right)} \right) = \lambda \nabla g\left( {x\left( c \right),y\left( c \right)} \right)\cdot\left( {x'\left( c \right),y'\left( c \right)} \right)$
But from part (a), we have $\nabla g\left( {x\left( c \right),y\left( c \right)} \right)\cdot\left( {x'\left( c \right),y'\left( c \right)} \right) = 1$. Therefore,
$\frac{d}{{dc}}f\left( {x\left( c \right),y\left( c \right)} \right) = \lambda $
c) In part (b) we obtain $\frac{d}{{dc}}f\left( {x\left( c \right),y\left( c \right)} \right) = \lambda $.
If we interpret $g\left( {x\left( c \right),y\left( c \right)} \right) = c$ as a "budget" constraint. Then $c$ is a "budget level". Since $f$ is a function of $c$, we conclude that $\lambda$ is the rate of increase in $f$ per unit increase in the "budget level" $c$.
