Three times a number t minus twelve equals forty
1 answer:
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Answer:
Not a right triangle.
Step-by-step explanation:
If this triangle is a right triangle, then the Pythagorean theorem would work on it. According to the Pythagorean theorem, , so c would always be greater than a or b. In this scenario, the greatest number here is 18, so c = 18. Since a or b don't matter if given both legs, we have the following equation that may or may not be equal:
. We know 8 squared is 64, and 11 squared is 121, and 18 squared is 324. This means that
according to the Pythagorean theorem, and since 64 + 121 = 185, and 185 does not equal 324, the Pythagorean theorem does not work and the triangle is not a right triangle.
Answer:
a)
Highest (-3,-3)
Lowest (2,2)
b)
Farthest (-3,-3)
Closest (2,2)
Step-by-step explanation:
To solve this problem we will be using Lagrange multipliers.
a)
Let us find out first the restriction, which is the projection of the intersection on the XY-plane.
From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:
completing the squares:
and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of
subject to the constraint
Now we have
Let be the Lagrange multiplier.
The maximum and minimum must occur at points where
that is,
we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so
Replacing in the constraint
from this we get
<em>x=-1/2 + 5/2 = 2 or x = -1/2 - 5/2 = -3 </em>
<em> </em>
and the candidates for maximum and minimum are (2,2) and (-3,-3).
Replacing these values in f, we see that
f(-3,-3) = 9+9 = 18 is the maximum and
f(2,2) = 4+4 = 8 is the minimum
b)
Since the square of the distance from any given point (x,y) on the paraboloid to (0,0) is f(x,y) itself, the maximum and minimum of the distance are reached at the points we just found.
We have then,
(-3,-3) is the farthest from the origin
(2,2) is the closest to the origin.