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.
3,000+4,000+1
Three hundred four thousand, one
Answer:
Simple interest, specifically in bank offers a grace period of six days, which gives time to payback. When saving money or investing, if the interest rate is high then the customer benefits. The disadvantages in simple interest are that if the interest rate is high then the borrower will pay more.
Answer:
Step-by-step explanation:
Since the results for the standardized test are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = test reults
µ = mean score
σ = standard deviation
From the information given,
µ = 1700 points
σ = 75 points
We want to the probability that a student will score more than 1700 points. This is expressed as
P(x > 1700) = 1 - P(x ≤ 1700)
For x = 1700,
z = (1700 - 1700)/75 = 0/75 = 0
Looking at the normal distribution table, the probability corresponding to the z score is 0.5
P(x > 1700) = 1 - 0.5 = 0.5