Optimization for Max Area Enclosed Rectangle on TI-89: Raw Transcript

This is a video from every step calculus dot com
demonstrating how my progams work on a t i eighty nine titanium calculator
and other calculators in the t i system for physics and calculus problems
ok this is an optimization problem in calculus and ah with regard to finding the area
given a amount of fence which is a usual problem in calculus
ah and let’s get started you put second alpha
you push second alpha to put in the i n d e x letters
and then press alpha and put in the eight and open and closed parenthesis
press enter and you’re into the menu calculus one menu
and we’re going to scroll down here to go to fence area
and we’re gonna press enter on there and a certain amount of
using a certain amount of fence how much is them maximum of area etcetera
given one side of a river and were going to put in maybe
you have to press alpha before you put any a numbers or any
characters in these lines of my programs so well press alpha
and were gonna put maybe eight fifty for the maximum fence that you have to work
with I always show you what you’ve entered
so you can change it if you want I say it’s ok
here’s a picture of it here’s the river
and you got the x on each side and the length of the other side
and generally in a problem or a test their going to ask you find the equation
for the length, here’s the equation eight fifty minus two x
write that on your paper here’s the area function
x times w and here’s the function here
write this on your paper eight fifty x minus two x squared
we multiplied x times eight fifty we take the derivative of that
here’s the derivative eight fifty minus four x
write that on your paper you gonna look like a genius
and we’re finding a critical number really critical number is not a critical point
because you haven’t found the y you’ve just found the x value
notice we took the derivative and then did the algebra computation
to find the x. write this on your paper
and here’s what you’ve found you found two thirteen on this side
two thirteen on this side and four twenty five for the length
just in case you’re interested in that they don’t generally ask you that
but they might you plug the critical number into the
original function to get the maximum area actually you’re getting the critical point
then and you do these computations and notice
that the area is ninety thousand three hundred thirteen
square units if its feet that would be square feet
if it’s meters what ever and this actually the point would be
eight fifty for the x and ninety thousand three hundred thirteen
for the y and pretty neat huh
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