Question

1. The art of being a good Chemical Engineer lies in being a good mathematician, a good chemist, and a good mechanic – all simultaneously. All three of these are required for determining optimum economics. Generally speaking, we concern ourselves with operating costs (i.e. the chemistry side) and capital costs (i.e. the mechanical side). What can decrease operating expenses will, most likely, increase capital expenses. When designing equipment, we MUST be able to reconcile these two. The sweet spot is usually at the intersection of two curves (i.e. the mathematician side). Suppose that Operating Expenses could be expressed by the equation: y = 0.5x + 0.15, and Capital Expenses could be expressed by y = 1.25x2 + 0.1. To determine the minimal costs, you must find the point of intersection (assuming that only positive roots are applicable). (Solve analytically and graphically.) NOTE: When solving graphically, use graphing paper. 2. As you will learn in your chemistry courses and the reactor design course, the rate at which a chemical reaction proceeds is very much dependent upon the reaction temperature.

Answer 1

Answer:

$y_{oe}=0.5x+0.15\\y_{ce}=1.25x^2+0.1$

for minimum cost the intersection point should be calculated i-e

$0.5x+0.15=1.25x^2+0.1\\\\1.25x^2-0.5x-0.5=0$

By using calculator

$x_1=0.4828\\x_2=-0.4828$

As x can't be negative so x=0.4828

It's the minimum value because as we decrease the operating cost further the capital value will increase so this is the minimum value.

Graphical solution: