If x^4+y^4=100, then the greatest possible value of x is between
A. 0 and 3
B. 3 and 6
C. 6 and 9
D. 9 and 12
E. 12 and 15

General rule for such kind of problems:
to maximize one quantity, minimize the others;
to minimize one quantity, maximize the others.

So, to maximize \(x\) we should minimize \(y^4\). Least value of \(y^4\) is zero. In this case \(x^4+0=100\) --> \(x^4=100\) --> \(x^2=10\) --> \(x=\sqrt{10}\approx{3.2}\), which is in the range (3,6).

Answer: B.
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\(x^4 + y^4 = 100\)
When you see even powers, first thing that should come to your mind is that the term will be positive or zero.
If you want to maximize x in the sum, you should minimize y^4 so that this term's contribution in 100 is minimum possible. Since it is an even power, its smallest value is 0 when y = 0.

Then \(x^4\) = 100
Since \(3^4 = 81\) and \(4^4 = 256\),x will lie between 3 and 4.
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Step 1:
x^4+y^4=100

x^4 will be maximum when y^4 is minimum. Lets assume y=0.1 so, y^4=0.0001

Step 2:

x^4+1^4=100
=> x^4+0.0001=100
=>x^4=100-0.0001
=>x^4=99.9999

Lets substitute x=3, i.e 3*3*3*3 = 81

So the value of x can be little more than 3 because 4*4*4*4=256

So the answer is option (B)

14. Min/Max Problems

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Two things that we must consider in order to solve this problem are:

a) We do not look for an integer

b) We do not look for a specific number but we want to see the number we are looking for in what range falls....e.x it is positive ot it is greater than 10.....in our example all the answers give range....

solution has been given by minimizing Y meaning Y=0

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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For y = 0, \(x^4 = 100\)

\(3^4 = 81 & 4^4 = 256\)

Answer = B

Hi All,

The answers to this question provide a great 'hint' as to how to go about solving it; since they're all essentially 'ranges', you can use them to figure out which solution contains the maximum value of X.

We're told that X^4 + Y^4 = 100. To maximize the value of X, we need to minimize the value of Y^4. The smallest that Y^4 could be is 0 (when Y = 0), so we'll have....

X^4 = 100

Looking at the answers, it makes sense to see what 3^4 equals....

3^4 = 81

Since that is BELOW 100, and 6^4 will clearly be MORE than 100, we have the correct answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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In an equation containing more than one variables, if we need to maximize one, it means that we have to minimize the others.
Since we have an even power of x and y, the minimum value of the variables can be 0

Therefore to maximize x, we need to minimize y.
Putting y = 0, x^4 = 100,
Or x = 3.xx

Hence Option C

The key to solving this problem for me was understanding that x is the maximum, when y is the minimum. Since y is raised to the fourth power, the smallest y can be is 0.

Here is the full solution:
\(x^4+y^4=100\)
\(x^4+0=100\)
\(x^4=100\)
\(x^2=10\)
\(x=\sqrt{10}\)
\(4>x>3\)

The correct answer is B

In determining the greatest possible value of x, we want to minimize y^4. Since the minimum value of y^4 is 0 (when y = 0), we have:

x^4 + 0 = 100

x^4 = 100

x^2 = 10

x = +/-√10

x ≈ 3.2 or -3.2

Thus, we see that the greatest value of x is between 3 and 6.

Answer: B
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Given: x^4 + y^4 = 100

Asked: The greatest possible value of x is between

For maximum x; y = 0
x^4 = 100
\(x = \sqrt{10} = 3.2\)

IMO B
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Whenever I see a^2 +b^2= c questions, I am reminded of th circle equation.

So in this case (x^2)^2 + (y^2)^2 =10^2

So since x^2 is a non negative value, I must lie between
0<=x^2<=10 .

So max(X) = √10 which lies between 3 and 4

Posted from my mobile device

A common mistake in such type of questions can be that one assumes that x and y are integers.
so y cannot be 0. by assuming this the option becomes A.
but it is nowhere given that they are integers. and cannot be equal to 0.
so we maximise x. which gives us x = 100^1/4 and y = 0.
we get x = 3.16