This week in calculus we worked on exam review packets. We have 6 of them, three are multiple choice and three are free response. I am getting much better at the multiple choice questions, especially when they ask you to find a limit when you are given a graph. I also understand how to find left and right hand limits better:
when they ask you to find the lefthand limit of 3. you cover up the right side of the graph and see when the line is approaching on the y axis on the left side. When they ask you to find the righthand limit of 3, you cover up the left hand side of the graph and see what y value the line is approaching on the right side. After finding those two, if you are asked to find the limit of 3. If the y values are not the same the limit does not exist, if they are the same then the limit does exist. Also by studying i have a better grasp on the first and second derivative test. Understanding those test better have helped me on a lot of the multiple choice questions i did not understand how to do earlier in the week. On the free response sheets, i understand how to do more limit problems. When they ask what limit is greater lim x approaches 1 or f(2). you look at the functions they give you. One is x=1 and the answer is 3. So by x=1 the limit is 3. Then there is a function 2 is less than or equal to x. So you plug f(2) into that function and the limit is x-1 which gives you 1. There fore as x approaches 1 is bigger than f(2).
The only thing i am still having problems with is some optimization problems. I understand now how to find primary and secondary functions but sometimes i get stuck in actually doing the problem. If someone can lists some easy steps on how to do optimization it would help a lot. Thanks :)
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ok for the optimization question you have...lets look at number four on the test we just got back.....[3.7 Homework Quiz .. out of 20 points]
ReplyDeleteso we know that we have a perimeter of 24 right...well to make your secondary and primary equation you make two squares...we know that the area of a square is lenth times width! and that there is two lengths and two widths for each square so your seconday equation will be 2(2L+2w)...that's your perimater because you're adding them...and since we know that the sum of the two perimeters is 24 you set 2(2L+2w) equal to 0!
so now...dealing with optimization:
take your secondary equation and solve for...lets go with length
2[2(L)+2(w)] = 0
24=4(L)+4(w)
24-4(w)=4(L)
(L)=6-w
so now plug in L to the primary equation which is your area...so that's
A=(L)(w)
A=(6-w)(w)
A=6w-(w)^2
TAKE THE DERIVATIVE OF THAT:
A'=6-2w
-2w=-6
w=3
SO NOW TAKE WHAT YOU KNOW....W=3....and PLUG IT INTO THE SECONDARY FORMULA:
24=2(2(L)+2(w)
24=2(2(L)+2(3)
24=4L+12
12=4L
L=3
so knowing your W is 3 and your L is 3
[it's a square so they had to be the same]
you know that the width is three and the length is three!!