The crop yield Y for a farm, in bushels, based on h inches of annual r

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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EDIT: the second term is essentially “-2p(h-mn)^2”, so set h=mn

­THANKS FOR EXTENDING MY APPROACH

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