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MT1302
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The crop yield Y for a farm, in bushels, based on h inches of annual r 30 Aug 2024, 05:20
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C
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Difficulty:

85% (hard)

Question Stats:

54% (02:51) correct 46% (02:40) wrong based on 147 sessions

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­The crop yield Y for a farm, in bushels, based on h inches of annual rainfall is modeled by the function \(Y(h) = 4p – 2p(h^2 – 2mnh + m^2n^2 ) + \frac{p}{(mn^2 )}- \frac{(mp^2)}{12}\), where m, n, and p are positive constants. What is the maximum crop yield?

(1) 12p – (mnp)^2 = 6mn^2

(2) p = 10­
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C
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pappal
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The crop yield Y for a farm, in bushels, based on h inches of annual r 03 Sep 2024, 14:22
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using differential dy/dh=0 we can find out the value of h for which the expression can be maximum
dy/dh=-4ph+4pmn=0 will give h=mn (since derivative of all constant values are always zero) is the value of h for which given expression will be maximum
putting h=mn in the expression and solving algebraically together with both the values mentioned in the statements 1 and 2 together will reduce the expression to a constant value 486/12 which is independent of any variable and hence the maximum value.
so C is the answer
BUT IMO THE QUESTION IS BEYOND THE SCOPE OF GMAT.­
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vijayakrishnan
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The crop yield Y for a farm, in bushels, based on h inches of annual r 03 Sep 2024, 20:42
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pappal
using differential dy/dh=0 we can find out the value of h for which the expression can be maximum
dy/dh=-4ph+4pmn=0 will give h=mn (since derivative of all constant values are always zero) is the value of h for which given expression will be maximum
putting h=mn in the expression and solving algebraically together with both the values mentioned in the statements 1 and 2 together will reduce the expression to a constant value 486/12 which is independent of any variable and hence the maximum value.
so C is the answer
BUT IMO THE QUESTION IS BEYOND THE SCOPE OF GMAT.­
Not necessarily. Combine the last two terms in the equation. You can then use the first statement to reduce the combined last two terms to “1/2”. Now we are only left with one negative term. That negative term is essentially “-2p(h+mn)^2. In order to maximize the value of Y, you should set this term to zero, which will happen when h=-mn. The only unknown is p, the value of which is given in the second statement

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vijayakrishnan
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The crop yield Y for a farm, in bushels, based on h inches of annual r 03 Sep 2024, 20:43
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vijayakrishnan
pappal
using differential dy/dh=0 we can find out the value of h for which the expression can be maximum
dy/dh=-4ph+4pmn=0 will give h=mn (since derivative of all constant values are always zero) is the value of h for which given expression will be maximum
putting h=mn in the expression and solving algebraically together with both the values mentioned in the statements 1 and 2 together will reduce the expression to a constant value 486/12 which is independent of any variable and hence the maximum value.
so C is the answer
BUT IMO THE QUESTION IS BEYOND THE SCOPE OF GMAT.­
Not necessarily. Combine the last two terms in the equation. You can then use the first statement to reduce the combined last two terms to “1/2”. Now we are only left with one negative term. That negative term is essentially “-2p(h+mn)^2. In order to maximize the value of Y, you should set this term to zero, which will happen when h=-mn. The only unknown is p, the value of which is given in the second statement

Posted from my mobile device

EDIT: the second term is essentially “-2p(h-mn)^2”, so set h=mn
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pappal
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The crop yield Y for a farm, in bushels, based on h inches of annual r 04 Sep 2024, 03:29
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vijayakrishnan
pappal
using differential dy/dh=0 we can find out the value of h for which the expression can be maximum
dy/dh=-4ph+4pmn=0 will give h=mn (since derivative of all constant values are always zero) is the value of h for which given expression will be maximum
putting h=mn in the expression and solving algebraically together with both the values mentioned in the statements 1 and 2 together will reduce the expression to a constant value 486/12 which is independent of any variable and hence the maximum value.
so C is the answer
BUT IMO THE QUESTION IS BEYOND THE SCOPE OF GMAT.­
Not necessarily. Combine the last two terms in the equation. You can then use the first statement to reduce the combined last two terms to “1/2”. Now we are only left with one negative term. That negative term is essentially “-2p(h+mn)^2. In order to maximize the value of Y, you should set this term to zero, which will happen when h=-mn. The only unknown is p, the value of which is given in the second statement

Posted from my mobile device
­THANKS FOR EXTENDING MY APPROACH
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SatvikVedala
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The crop yield Y for a farm, in bushels, based on h inches of annual r 22 Sep 2024, 07:41
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MT1302
­The crop yield Y for a farm, in bushels, based on h inches of annual rainfall is modeled by the function \(Y(h) = 4p – 2p(h^2 – 2mnh + m^2n^2 ) + \frac{p}{(mn^2 )}- \frac{(mp^2)}{12}\), where m, n, and p are positive constants. What is the maximum crop yield?

(1) 12p – (mnp)^2 = 6mn^2

(2) p = 10­
Given \(Y(h) = 4p – 2p(h^2 – 2mnh + m^2n^2 ) + \frac{p}{(mn^2 )}- \frac{(mp^2)}{12}\)

To find maximum of Y(h) find dY(h)/dh & make it zero

Y(h) = \(Y(h) = 4p – 2p(h-mn)^2 + \frac{p}{(mn^2 )}- \frac{(mp^2)}{12}\)

Y'(h) = -2p(2(h-mn))

0 = -4p(h-mn)

h = mn --------------- (I)

Substituting I in Y(h) for maximum value

Y(h) = \(Y(h) = 4p – 2p(mn-mn)^2 + \frac{p}{(mn^2 )}- \frac{(mp^2)}{12}\)

Y(h) = \(Y(h) = 4p – \frac{p}{(mn^2 )}- \frac{(mp^2)}{12}\)

Solving Y(h) gives [48*p*m*n^2+12p-(mnp)^2]/12*m*n^2

From Statement 1 we have 12p-(mnp)^2 = 6mn^2

Y(h) = [48p+6] ------- not sufficient

From Statement 1 we have just p = 10 which when considered alone is not sufficient

from i & ii

Y(h) becomes = 486/12
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The crop yield Y for a farm, in bushels, based on h inches of annual r 31 May 2025, 23:22
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Can avoid using calculus; this is a gmat qs. Attaching solution.

Side note - In gmat, in most of the min-max qs, you will have a sq term that can be set to 0.
MT1302
­The crop yield Y for a farm, in bushels, based on h inches of annual rainfall is modeled by the function \(Y(h) = 4p – 2p(h^2 – 2mnh + m^2n^2 ) + \frac{p}{(mn^2 )}- \frac{(mp^2)}{12}\), where m, n, and p are positive constants. What is the maximum crop yield?

(1) 12p – (mnp)^2 = 6mn^2

(2) p = 10­

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