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how do you solve this?

Maximize (A 3)(B-2) if A B=40

$(A+3)(B-2),A+B=40$

Substitution: $A=40-B\Longrightarrow (A+3)(B-2)=(43-B)(B-3)=-B^2+46B-129$

Now differentiate the expression, and impose it equal to 0: $\frac{d}{dB}(-B^2+46B-129)=-2B+46=0$
which leads to B=23. Second derivative is negative, which means B=23 is a local maximum. Substitute B=23 in the original expression.

Oops, there is a typo with B-3 instead of B-2:

$(43-B)(B-2)= -B^2 + 45B -46$

using the same argument you have B=45/2 which maximizes the expression.

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