How to isolate a variable inside floor and ceiling functions?

algebra-precalculus ceiling-and-floor-functions

Solution 1:

You cannot solve for $E_c$ in general, but you can determine a range of potential solutions using the inequalities that define the floor/ceiling functions $\,\lceil x \rceil - 1 \lt x \le \lceil x \rceil\,$ and $\,\lfloor x \rfloor \le x \lt \lfloor x \rfloor + 1\,$:

$$ F_c - 1 \;\lt\; Y + \frac{D_3 - Z \cdot Y}{2 \cdot Z} = \frac{1}{2}\left(\left\lceil\frac{D_1}{Z}\right\rceil + \frac{D_3}{Z}\right) \;\le\; F_c \\ 2 F_c - \frac{D_1}{Z} - 3\;\lt\; 2F_c - \left\lceil\frac{D_1}{Z}\right\rceil - 2 \;\lt\; \frac{D_3}{Z} \;\le\; 2F_c - \left\lceil\frac{D_1}{Z}\right\rceil \;\le\; 2F_c - \frac{D_1}{Z} \\ 2F_c - 3 \;\lt\; \frac{D_1+D_3}{Z} \;\le\; 2F_c \\ \frac{D_1+D_3}{2F_c} \;\le\; Z = \left\lfloor\frac{A \cdot E_c}{100}\right\rfloor \;\lt\; \frac{D_1+D_3}{2F_c - 3} \\ \left\lceil \frac{D_1+D_3}{2F_c} \right\rceil \;\le\; \frac{A \cdot E_c}{100} \;\lt\; \left\lceil \frac{D_1+D_3}{2F_c - 3} \right\rceil \\ \frac{100}{A}\left\lceil \frac{D_1+D_3}{2F_c} \right\rceil \;\le\; E_c \;\lt\; \frac{100}{A}\left\lceil \frac{D_1+D_3}{2F_c - 3} \right\rceil $$

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