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.
Is there a graph or chart? Or a word problem that tells us some information
Answer:
-499,485.
Step-by-step explanation:
We can transform this to an arithmetic series by working it out in pairs:
6^2 - 7^2 = (6-7)(6+7) = -13
8^2 - 9^2 = (8-9)*8+9) = -17
10^2 - 11^2 = -1 * 21 = -21 and so on
The common difference is -4.
The number of terms in this series is (998 - 6) / 2 + 1
= 992/2 + 1 = 497.
Sum of n terms of an A.S:
= n/2 [2a1 + (n - 1)d
Here a1 = -13, n = 497, d = -4:
Sum = (497/2)[-26 - 4(497-1)]
= 497/2 * -2010
= -499,485.
A = x * x = x^2
B = 6 * x = 6x
C = 8 * x = 8x
D = 6 * 8 = 48
So the area of the whole shape is each side added up then multiplied to give (x+6)(x+8) =
x^2 + 14x + 48