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.
It reveals the slope and a single point on the line
Question 3. The true statements are:
4g2 – g = g2(4 – g) ⇒ should be: 4g² - g = g(4g - 1)
9g3 + 12 = 3(3g3 + 4) ⇒ should be: 9g³ + 12 = 3(3g³ + 4) TRUE
24g4 + 18g2 = 6g2(4g2 + 3g) ⇒ should be: 24g⁴ + 18g² = 6g²(4g² + 3)
<span>35g5 – 25g2 = 5g2(7g3 – 5) </span>⇒ should be: 35g⁵ - 25g² = 5g²(7g³ - 5) TRUE
Question 4. Completely factored.
16y⁵ + 12y³ = 4y³(4y² + 3) FACTORED COMPLETELY
18y³ - 6y = 6y(3y² - 1)
20y⁷ + 10y² = 10y²(2y⁵ + 1)
32y¹⁰ - 24 = 8(4y¹⁰ - 3) FACTORED COMPLETELY