The rate (in mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function p = 120i i2 + i + 9 where i is the light intensity (measured in thousands of foot-candles). for what light intensity is p a maximum?

The given function is:

P = 120 i / (i^2 + i + 9)

or

P = 120 i (i^2 + i + 9)^-1

The maxima point is obtained by taking the 1st derivative of the function then equating dP / di = 0:

dP / di = 120 (i^2 + i + 9)^-1 + (-1) 120 i (i^2 + i + 9)^-2 (2i + 1)

setting dP / di =0 and multiplying whole equation by (i^2 + i + 9)^2:

0 = 120 (i^2 + i + 9) – 120i (2i + 1)

Dividing further by 120 will yield:

i^2 + i + 9 – 2i^2 – i = 0

-i^2 + 9 =0

i^2 = 9

i = 3      (ANSWER)

Therefore P is a maximum when i = 3

Checking:

P = 120 * 3 / (3^2 + 3 + 9)

P = 17.14

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g