On the bounds of the objective function in a standard LP Consider a standard linear programming (LP) such as:
\begin{align} \sum_{i=1}^{N}\frac{a_{i}}{b_{i}}x_{i}\end{align}  
\begin{align}\text{s.t. }\left ( \sum_{i=1}^{N}x_{i}=1 \; , \; \sum_{i=1}^{N}b_{i}x_{i}=c>0\right )\end{align}
Note:
The variables and coefficients are positive.
Can we determine lower- and upper-bounds for the objective function, in terms of $a_{i},b_{i}, c$?
 A: In general, the simple expressions, in terms of $a_i,b_i,c$, for the required upper and lower bounds are $\max\left\{ \dfrac{a_i}{b_i} : 1 \le i \le N \right\}$ and $\min\left\{ \dfrac{a_i}{b_i} : 1 \le i \le N \right\}$ respectively.
As a consequence of the Fundamental Theorem of Linear Programming, I'll just consider the basic solutions of this LPP.
Suppose that $x_i,x_j$ (with $1 \le i < j \le N$ for convenience) are chosen as the basic variables.  Then we have the basis matrix
\begin{equation*}
  B = \begin{bmatrix}
    1 & 1 \\
    b_i & b_j
  \end{bmatrix},
  x_B = \begin{bmatrix}x_i \\ x_j\end{bmatrix},
  b = \begin{bmatrix}1 \\ c\end{bmatrix}.
\end{equation*}
$\therefore Bx_B = b$.  If $x_i,x_j \ge 0$, from the first constraint $x_i + x_j = 1$, we know that the value of the objective function is the convex combination of $\dfrac{a_i}{b_i}$ and $\dfrac{a_j}{b_j}$.
\begin{align*}
  & \quad\sum_{i=1}^{N}\frac{a_{i}}{b_{i}}x_{i} \\
  & = \frac{a_i}{b_i} x_i + \frac{a_j}{b_j} x_j \\
  & = \frac{a_i}{b_i} x_i + \frac{a_j}{b_j} (1 - x_j) \\
  & \in \left[ \min\left\{ \frac{a_i}{b_i},\frac{a_j}{b_j} \right\},\max\left\{ \frac{a_i}{b_i},\frac{a_j}{b_j} \right\} \right]
\end{align*}
Hence, the problem is done.  I include more conditions for the feasibility of $x_B = (x_i,x_j)^T$.
Case 1: $b_i = b_j$
If the basic solution is feasible, then $b_i = b_j = c$ and $x_B = (0,1)^T$ or $(1,0)^T$.
Case 2: $b_i \ne b_j$
Then $B$ is invertible.
\begin{equation*}
  B^{-1} = \frac{1}{b_j - b_i} \begin{bmatrix}
    b_j & -1 \\
    -b_i & 1
  \end{bmatrix}
\end{equation*}
Compute $x_B = B^{-1} b$.  Note that it should be non-negative.
\begin{equation*}
  x_B = \begin{bmatrix}x_i \\ x_j\end{bmatrix}
  = \frac{1}{b_j - b_i} \begin{bmatrix}
    b_j & -1 \\
    -b_i & 1
  \end{bmatrix}
  \begin{bmatrix}1 \\ c\end{bmatrix}
  = \begin{bmatrix}
    \dfrac{b_j - c}{b_j - b_i} \\ \dfrac{c - b_i}{b_j - b_i}
  \end{bmatrix} \ge 0
\end{equation*}
If you assume either $b_j > b_i$ or $b_i > b_j$, then we have $b_i \le c \le b_j$ or $b_j \le c \le b_i$.
