Problem+10+Deborah



Trying to Maximize volume V=LxBxH the height is what changes the length and base in order for it to match up at the top, h is the only variable. squares of length h are cut out, and the remainder is a b and a l

5 sides short box with large base tall box with small base

original length is 3ft 3ft=2h+b V=bhl b=3-2h square piece of cardboard 3 ft long=3 ft high 3ft=2h+l l=3-2h V=(3-2h)^2 h This is the objective function!

V'=(3-2h)^2(1) + h(2(3-2h)*-2) (ms. sweeney) I disagree here -- you are missing a negative! Oops, got it! V'=(3-2h)^2 - 4h(3-2h) 0=9-12h+12h^2; h=3/2 or 1/2 h must be less than 3 (because the original dimensions of the cardboard are 3 by 3) and greater than 0 (because if there was nothing cut out, the box would have no sides). When 3/2 is plugged in, it gives a volume of 0, so the answer is h = 1/2 foot. Is this right? It's weird that 3/2 doesn't work even though it's within the domain so I'm not sure. (Maybe I factored wrong again?)