Job Search and the Investment Problem I don't know what to take as the cost of accepting the job. Kindly guide me through deciding how to frame the equation.
You interview one employer per week, and your discount rate is $\delta = 1/(1+.07)$.  Suppose $x$ is the lowest quality job you would accept: anything worse than $x$, and you reject and keep searching. Let $J$ be the value of searching. Then we have the functional equation:
$$
J = \max_{x} \int_{x}^{1}z dF(z) + \int_{0}^x \delta J dF(z)
$$
The first term is the jobs you accept, since they are above your cutoff, and the second are the scenarios where you keep searching and forego the bad offer, so that you end up where you started tomorrow with an expected payoff of $J$. You have a uniform distribution so $dF(z) = dz$ on $[0,1]$.

Maximizing over $x$ gives the optimization condition
$$
-x*f(x*) + \delta f(x*) \delta J = 0
$$
so that $x^* = \delta J$.  Only if the offer is better than $\delta J$, the discounted value of continuing, do you accept; indeed, $\delta J$ is the opportunity cost of accepting, so I assume that is your $c$.

Now, let's solve for $J$. We have the equation
$$
J = \int_{\delta J}^1 zdz + \int_0^{\delta J}dz \delta J
$$
$$
J = \dfrac{1}{2}(1-(\delta J)^2) + (\delta J)^2
$$
$$
J = \dfrac{1+(\delta J)^2}{2}
$$
$$
0 = 1+\delta^2 J^2 - 2J
$$
This is a quadratic equation. The solution is 
$$
J^* = \dfrac{1}{\delta^2} - \dfrac{\sqrt{4 - 4\delta^2}}{2\delta^2}
$$
$$
J^* = \dfrac{1-\sqrt{1 - \delta^2}}{\delta^2}
$$
and your search strategy is to accept if the job is better than 
$$
x^* = \dfrac{1-\sqrt{1 - \delta^2}}{\delta}
$$
and reject otherwise.