It would approach infinity. You can make x negative and y positive which would make the expression positive. Remember the fifth power of infinity is infinity. Also infinity does not exist it is only a theory so the number cannot be infinity it would approach infinity.
I like how I posted the correct answer as my first answer yet others decided to chime in with their responses, most of which were incorrect anyway. Then I get a thumbs down as the only one who gets feedback. Blasphemy, if you are retarded then you can read the solution below by alph but if you truly want to understand mathematics as I certainly already do you must gain some intuition. Those referring to the local maxs/mins obviously have no mathematical insight and rely on memorizing methods which they have no idea what they are used for anyway. Please do not comment on something which you know nothing about and I find the infinity to the fifth power to be quite a humorous anecdote.
One point shall be given to alph, two to me, and none to the rest of this discussion especially the one who posts there must be more to this question after the correct answer is indeed given. But possibly he might be referring to how someone cannot realize the answer to this question from meer inspection. Is it possible that you are asking x to the 2y power to the 3rd power.
2007-01-14 20:20:13
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answer #1
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answered by Anonymous
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3x + 4y = 5
Rearrange to find x :
x = (5 - 4y) / 3
Now, x^2 * y^3 = [(5 - 4y) / 3]^2 * y^3
Now take the derivative and set it equal to zero.
[(5 - 4y) / 3]^2 * 3y^2 + 2 * [(5 - 4y) / 3] * (-4 / 3) * y^3 = 0
Gathering up some terms and rearranging, we have :
3y^2 * [(5 - 4y) / 3]^2 = 8y^3 * [(5 - 4y) / 3] / 3 (= Equation 1)
Now if y = 0, then x^2 * y^3 = 0, which is not much of a maximum.
Also, if (5 - 4y) / 3 = 0, implying that y = 5 / 4, then in the
equation, 3x + 4y = 5, this implies that x = 0, and again
we are left with x^2 * y^3 = 0.
What this means, is that, if we are prepared to accept that the
maximum will not be zero, then we can cancel out y^2 and
(5 - 4y) / 3 from Equation 1, because neither term is zero.
So Equation 1 becomes :
3(5 - 4y) / 3 = 8y / 3
From this, we obtain y = 3/4.
Substituting into 3x + 4y = 5 gives : x = 2/3
The greatest value of x^2 * y^3 is therefore :
(2/3)^2 * (3/4)^3 = 3/16.
Edit : provided x and y are both greater than zero.
2007-01-14 21:41:38
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answer #2
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answered by falzoon 7
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Since 3x + 4y = 5, it follows that
3x = 5 - 4y
x = (5 - 4y)/3
Let G = (x^2)(y^3). Then, making the substitution x = (5 - 4y)/3,
G = [(5 - 4y)/3]^2 (y^3)
Expanding this,
G = (1/9)[(5 - 4y)^2] (y^3)
Expanding the squared binomial,
G = (1/9) [25 - 40y + 16y^2][y^3]
Distributing the y^3,
G = (1/9) [25y^3 - 40y^4 + 16y^5]
So this is the function we want to maximize.
G(y) = (1/9) [25y^3 - 40y^4 + 16y^5]
To find the maximum value, we find the derivative G'(y), and then make it 0.
G'(y) = (1/9) [75y^2 - 160y^3 + 80y^4]
Setting G'(y) = 0,
0 = (1/9) [75y^2 - 160y^3 + 80y^4]
Dividing both sides by 1/9,
0 = [75y^2 - 160y^3 + 80y^4]
And then dividing both sides by 5,
0 = 15y^2 - 32y^3 + 16y^4
0 = 16y^4 - 32y^3 + 15y^2
0 = y^2 (16y^2 - 32y + 15)
This factors as follows:
0 = y^2 (4y - 3) (4y - 5)
Solving for y gives us our critical values,
y = {0, 3/4, 5/4}
Now that we have our critical values, we need to determine the intervals of increase. Recall that
G'(y) = (1/9) [75y^2 - 160y^3 + 80y^4]
G'(y) = (5/9) (y^2) (4y - 3) (4y - 5)
And we can easily determine the absolute max.
Test -1: (5/9) (positive) (negative) (negative) = positive.
Therefore, G(y) is increasing on (-infinity, 0].
Test 1/2: (5/9) (positive) (negative) (negative) = positive.
G(y) is increasing on [0, 3/4]
Test 4/4, or 1: (5/9) (positive) (positive) (negative) = negative.
Therefore, G(y) is DECREASING on [3/4, 5/4]
Test 2: (5/9) (positive) (positive) (positive) = positive.
G(y) is increasing on [5/4, infinity)
G(y) is increasing on (-infinity, 3/4] U [5/4, infinity]
G(y) is decreasing on [3/4, 5/4]
That means there is a local maximum at y = 3/4, and a local minimum at 5/4.
To obtain the maximum value, plug in G(3/4)
G(3/4) = (1/9) [25(3/4)^3 - 40(3/4)^4 + 16(3/4)^5]
G(3/4) = (1/9) (3/4)^3 [25 - 40(3/4) + 16(3/4)^2]
G(3/4) = (1/9) (3/4)^3 [25 - 30 + 16(9)]
G(3/4) = (1/9) (3/4)^3 [-5 + 144]
G(3/4) = (1/9) (27/64) [139]
G(3/4) = 417/64
2007-01-14 20:34:55
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answer #3
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answered by Puggy 7
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Take 3x+4y=5 and solve for x in terms of y:
3x+4y=5
3x=5-4y
x=(1/3)(5-4y)
Substitute this into (x^2)(y^3):
[(1/3)(5-4y)]^2(y^3)
Simplify to get:
(1/9)(25-40y+16y^2)(y^3)
(25/9)y^3 - (40/9)y^4 + (16/9)y^5
Notice that the highest degree in this expression is 5.
This means that the term (16/9)y^5 will dominate and so the greatest value of the entire expression will certainly depend on this term. But, plugging in infinitely large values into the quintic will result in even larger values.
Conclusion: There is no "greatest" value.
Edit:
Be careful about using first/second derivative methods to determine greatest values. They only give you LOCAL max/min values. If this is indeed what you are looking for, then refer to Puggy's answer (see below). But, if you are looking for the greatest possible value (i.e. absolute max/min), then such tests won't work on odd degree polynomials.
2007-01-14 20:31:02
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answer #4
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answered by alsh 3
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Rewrite - -5 < (3x - 2) < 5 to discover the set, you may resolve for all of the two inequalities. the respond may be the intersection of the two instruments. you may shortcut to the right value of x by utilising merely doing the righthamd ingredient - 3x-2<5 == 3x< 7 x<7/3, so the right interger x could be may be 6/3 = 2
2016-12-16 05:03:25
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answer #5
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answered by ? 4
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rearrange 3x + 4y = 5
x = (5 - 4y)/3
now plug x into (x^2)(y^3) <-- let's make it g(y)
(((5 - 4y)/3)^2)(y^3)
take the derivative of g(y)
then make g'(y) = 0
then solve for y
and do the 2nd derivative test
which you take the 2nd derivative of g(y)
and plug-in the y value/s you get from g'(y) = 0
in order to find the greatest value of (x^2)(y^3)
you need the y value that when you plug it into g"(y), it makes g"(y) negative
then you use that y value to solve for x
by plug it into 3x + 4y = 5
2007-01-14 20:51:33
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answer #6
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answered by ? 2
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x can be negative, which lets y be more positive. As x approaches negative infinity, y approaches positive infinity. So the greatest value of x squared times y cubed is the fifth power of infinity. It has too many digits to include in this message.
2007-01-14 20:21:34
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answer #7
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answered by x4294967296 6
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+ or - infinity except for x = 0
2007-01-14 20:32:58
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answer #8
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answered by Anonymous
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I have to believe this question is lacking some information.
2007-01-14 20:29:09
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answer #9
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answered by gebobs 6
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