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 Originally Posted by miracleriver
Originally Posted by martindcx1e
so x^2 + 4xh = 1200.
h = (1200 - x^2)/(4x)
Volume = (x^2) * [(1200 - x^2)/(4x)]
Which = 300x - (1/4)x^3
Deriv of that = 300 - (3/4)x^2
x = 20
(20)^2 + (4)(20)(h) = 1200
h = 10
Ok so you find the equation for what you want to minimize or maximize. Then you find another equation to let you isolate one variable. Then substitute into the original equation, take the derivative, set it to 0, and solve for that variable. Then plug that answer back into your 2nd equation to find the other variable. Is that pretty much it?
For full credit you probably need to show that it is a maximum (as opposed to a minimum or an infliction point)
Depends on the instructor. You're right obv but in my experience most instructors don't require it for entry-level Calculus students.
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