Lagrange multipliers:







(if

)

(if

)

(if

)
In the first octant, we assume

, so we can ignore the caveats above. Now,

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of

.
We also need to check the boundary of the region, i.e. the intersection of

with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force

, so the point we found is the only extremum.
2/33 * 1/5 * 11/10 = (2 * 1 * 11) / (33 * 5 * 10) = 22/1650 reduces to 1/75
The length of the envelope is 6.3 inches
Step-by-step explanation:
Width of the envelope = 3 inches
Diagonal of the envelope = 7 inches
To find:
The length of the envelope.
Let the length of the envelope be 'l'
We can use pythogoras theorem to calculate the length.
l = √(49-9)
l = √(40)
l = 6.3 inches
The length of the envelope is 6.3 inches
1 electron has a mass of 9.1 X 10-31 kg. How many electrons n does it take to make 1 kg?
(1 e)/(9.1 X 10-31 kg) = n/(1 kg)
So, n = 1.10 X 1030 electrons
Each electron has a charge of 1.6 X 10-19 C, where C stands for Coulombs. So, the n electrons have a charge of q = ne = (1.10 X 1030 electrons) X (1.6 X 10-19 C/electron) = 1.76 X 1011 C.