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avggmatstudent23
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If -1/2<x<-1/9 and -1/4<y<1/25, what is the maximum value of xy^2 29 Jan 2023, 06:58
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If \(\frac{-1}{2}\leq x \leq\frac{-1}{9}\) and \(\frac{-1}{4}\leq y \leq\frac{1}{25}\), what is the maximum value of \(xy^2\)

(A) \(\frac{-1}{16}\)
(B) \(\frac{-1}{32}\)
(C) \(\frac{-1}{144}\)
(D) \(\frac{-1}{5625}\)
(E) 0

I get the correct answer when I first square the y inequality and then multiply it by the x inequality i.e. \(y^2 * x\). However, I get the wrong answer when I first multiply the x and y inequalities and then multiply the y inequality again i.e. \(xy * y\).

What am I missing here?

Bunuel chetan2u


Source: GMATClub video with Crackverbal (see at 50:00 mark)
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gasoline
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If -1/2<x<-1/9 and -1/4<y<1/25, what is the maximum value of xy^2 29 Jan 2023, 10:54
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avggmatstudent23
If \(\frac{-1}{2}\leq x \leq\frac{-1}{9}\) and \(\frac{-1}{4}\leq y \leq\frac{1}{25}\), what is the maximum value of \(xy^2\)

(A) \(\frac{-1}{16}\)
(B) \(\frac{-1}{32}\)
(C) \(\frac{-1}{144}\)
(D) \(\frac{-1}{5625}\)
(E) 0

I get the correct answer when I first square the y inequality and then multiply it by the x inequality i.e. \(y^2 * x\). However, I get the wrong answer when I first multiply the x and y inequalities and then multiply the y inequality again i.e. \(xy * y\).

What am I missing here?

Bunuel chetan2u


Source: GMATClub video with Crackverbal (see at 50:00 mark)
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Hello

I hope you're well.
Do not stress. I believe that you're simply overthinking your approach.
Multiplying any three values should give you the same result; you may perform the operation in any order.

Options A till D are all negative, and only option E is 0 (non-negative).
0 is the largest of all the options.

The range of X does not cross the number line from negative to positive; however, the range of Y does so.

X will always be negative.
Y^2 will always be non-negative.

Multiplying X and Y^2, you will always get a negative output except when Y is 0.
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If -1/2<x<-1/9 and -1/4<y<1/25, what is the maximum value of xy^2 30 Jan 2023, 08:18
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gasoline
avggmatstudent23
If \(\frac{-1}{2}\leq x \leq\frac{-1}{9}\) and \(\frac{-1}{4}\leq y \leq\frac{1}{25}\), what is the maximum value of \(xy^2\)

(A) \(\frac{-1}{16}\)
(B) \(\frac{-1}{32}\)
(C) \(\frac{-1}{144}\)
(D) \(\frac{-1}{5625}\)
(E) 0

I get the correct answer when I first square the y inequality and then multiply it by the x inequality i.e. \(y^2 * x\). However, I get the wrong answer when I first multiply the x and y inequalities and then multiply the y inequality again i.e. \(xy * y\).

What am I missing here?

Bunuel chetan2u


Source: GMATClub video with Crackverbal (see at 50:00 mark)


Hello

I hope you're well.
Do not stress. I believe that you're simply overthinking your approach.
Multiplying any three values should give you the same result; you may perform the operation in any order.

Options A till D are all negative, and only option E is 0 (non-negative).
0 is the largest of all the options.

The range of X does not cross the number line from negative to positive; however, the range of Y does so.

X will always be negative.
Y^2 will always be non-negative.

Multiplying X and Y^2, you will always get a negative output except when Y is 0.
Show more

I already know that I get the right answer by multiplying x with y^2. My question is why don't I get the correct answer if I multiple xy with y.

Why don't you try solving it this way and let me know if you figure out where I'm going wrong.

Thanks!
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If -1/2<x<-1/9 and -1/4<y<1/25, what is the maximum value of xy^2 30 Jan 2023, 10:42
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If \(\frac{-1}{2}\leq x \leq\frac{-1}{9}\) and \(\frac{-1}{4}\leq y \leq\frac{1}{25}\), what is the maximum value of \(xy^2\)

(A) \(\frac{-1}{16}\)
(B) \(\frac{-1}{32}\)
(C) \(\frac{-1}{144}\)
(D) \(\frac{-1}{5625}\)
(E) 0

I get the correct answer when I first square the y inequality and then multiply it by the x inequality i.e. \(y^2 * x\). However, I get the wrong answer when I first multiply the x and y inequalities and then multiply the y inequality again i.e. \(xy * y\).

What am I missing here?

Bunuel chetan2u


Source: GMATClub video with Crackverbal (see at 50:00 mark)


Hello

I hope you're well.
Do not stress. I believe that you're simply overthinking your approach.
Multiplying any three values should give you the same result; you may perform the operation in any order.

Options A till D are all negative, and only option E is 0 (non-negative).
0 is the largest of all the options.

The range of X does not cross the number line from negative to positive; however, the range of Y does so.

X will always be negative.
Y^2 will always be non-negative.

Multiplying X and Y^2, you will always get a negative output except when Y is 0.

I already know that I get the right answer by multiplying x with y^2. My question is why don't I get the correct answer if I multiple xy with y.

Why don't you try solving it this way and let me know if you figure out where I'm going wrong.

Thanks!
Show more


Hello


I hope you're using the same values of Y when you're performing (X*Y)*Y.
Let us try to maximize XY first.
-1/2 * -1/4 = 1/8

If you use the same value of Y, then you end up getting a negative number, and I hope you're not using different values of Y when performing the two operations.


    negative x* negative y* negative y=negative
    negative x* positive y* positive y=negative
    negative x*0*0=0
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Adarsh_24
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If -1/2<x<-1/9 and -1/4<y<1/25, what is the maximum value of xy^2 21 Sep 2024, 03:16
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avggmatstudent23
If \(\frac{-1}{2}\leq x \leq\frac{-1}{9}\) and \(\frac{-1}{4}\leq y \leq\frac{1}{25}\), what is the maximum value of \(xy^2\)

(A) \(\frac{-1}{16}\)
(B) \(\frac{-1}{32}\)
(C) \(\frac{-1}{144}\)
(D) \(\frac{-1}{5625}\)
(E) 0

I get the correct answer when I first square the y inequality and then multiply it by the x inequality i.e. \(y^2 * x\). However, I get the wrong answer when I first multiply the x and y inequalities and then multiply the y inequality again i.e. \(xy * y\).

What am I missing here?

Bunuel chetan2u


Source: GMATClub video with Crackverbal (see at 50:00 mark)
Show more
I know exactly why you’re getting wrong answer.

I think you found extreme values of x*y first.
You might have found -1/50 and 1/8 as extremes. Then y has -1/4 and 1/25 as extreme. So you may have gotten 1/200.
But, the thing you might have missed is y has to be constant.
Let’s say xy had extreme 1/8 with x=-1/2 and y=-1/4. Then y has to be same -1/4 when you multiply xy*y.
So it is better to square y first and then multiply to not get bogged down by these complications.
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If -1/2<x<-1/9 and -1/4<y<1/25, what is the maximum value of xy^2 09 Jul 2025, 00:42
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Why is the ANS not D. If we do x*y*y then i am getting D as the Ans. Where am I going wrong in the current method or approach?
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If -1/2<x<-1/9 and -1/4<y<1/25, what is the maximum value of xy^2 09 Jul 2025, 12:50
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viswanath1
Why is the ANS not D. If we do x*y*y then i am getting D as the Ans. Where am I going wrong in the current method or approach?
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We want the maximum value of \(xy^2\)

We know that \(x\) is going to be negative, and \(y^2\) is going to be positive or 0.

So, \(xy^2\) can either be negative or 0

Since, negative < 0 and we want maximum value, so max. \(xy^2 = 0\)
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