Question

maximize the area of a triangle given the constraint b+h=32

Answer 1

This is calculus question, right?

\(A = \dfrac{1}{2}bh\) and \(b+h=32\)

\(h = 32-b\)

So we can make substitute here; \(A = \dfrac{1}{2}b(32-b)=-\dfrac{1}{2}b^2+16b\)

So you can find what value of b is when area of triangle is maximum. Doing that by derivative area in respect to b then set it equal to 0.

\(\dfrac{dA}{db} = -b+16 = 0\)

So we can see that \(b = 16\)

Now we can solve for h: \(b+h=32~~~\Longrightarrow~~~16+h=32\)

\(h = 16\)

So we have \(A = \dfrac{1}{2}(16)(16) =\boxed{128} \)