You are highly encouraged to work on the following problems this weekend in preparation for Monday's class and the Unit Test which is being written on Tuesday next week.
Suggested Review Problems for Ch. 3: Pages 156 - 159
#'s 1, 2, 3, 4, 5, 13, 28 and 30 cover material based on section 3.1
#'s 6, 7, 8 and 26 are concepts from section 3.2
#'s 9, 14, 15, 17, 18, 19, 20, 21 and 23 are the optimization problems from sections 3.3/3.4.
I promise to check the blog on a regular basis this weekend. If you post any questions I will certainly get back to you over the course of the day.
Have a great weekend! See everyone on Monday afternoon for REVIEW!!!!!
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Hi sir, I was wondering if you could help me with #14 on pg.157
ReplyDeleteA couple of obvious points: V=lwh and
ReplyDeleteSA= lw + 2wh + 2 lw (NOTE: only one lw since there is no top for this box)
Since they said the box has a square base we can assume l=w and just simply call it a box that has dimensions x by x by h.
So, the volume formula now becomes V=x^2h and the surface area formula will simplify to
SA= x^2 + 4xh
You can now substitute the 10,000 into the volume formula and isolate for h. This will give h= 10,000 / x^2.
Sub h into the 'new' SA formula for the square box to get SA= x^2 + 4x*(10,000/x^2)
Simplify the above formula and then take the derivative of SA with respect to x. Set the result to zero and solve for x. That should give the dimensions of the base which you can than use to find the height.
Let me know if you need any further guidance or if that was enough to get you started.
Mr. A.
CORRECTION: I just noticed that I made a typo in the first SA formula.
ReplyDeleteIt should have read
SA= lw + 2wh + 2lh Now there is only ONE lw since there is no top.
Sorry for any confusion this may have caused.
Thanks sir, I already solved that question. However I have a question for #15.
ReplyDeleteI solved it using the 2 formulas A=LW and
P=2L+ 2W. Since the area is already given I subbed it back into the equation and isolated for W. I got W=2400/l. I subbed that into the P. equation. However when I solve for L, I keep on getting the wrong answer.
Are the formulas I am using correct? Because I know my calculations are accurate.
hey sir question 28e in page 157 i didnt get the right answer as the back of the book i got -1/3
ReplyDelete@ Habib
ReplyDeletef(x)=(x+5)^1/2
Find the derivative
f'(x)= (1/2)(x+5)^-1/2
Find f"(x)
f"(x)=(-1/4)(x+5)^-3/2
f"(x) also equals to
(-1)/(4(x+5)^3/2)
Then you sub in x=4
oh ya i see were i went wrong
ReplyDeletesir if you have time question 30 i got the velocity but my acceleration is not the same as the back of the book
ReplyDeleteHabib,
ReplyDeleteThe answer for acceleration in the back of the book for question 30 is correct. After you apply the product rule to differentiate v(t) you get an expression (yes it's ugly looking) that can be simplified by pulling out the common factor. That common factor is
[t^2+t]^(-4/3).
When you simplify what is left in the bracket (you will have to find a common denominator of 3) eventually you will be able to factor out a common term of one third.
In doing so you will generate the simplified version of a(t).
See me tomorrow morning as it will take too long to show all of the work here...the trick is the common factor with the exponent of negative four thirds.
In regards to Question #15 from the review...
ReplyDeleteYou have to visualize what five (5) adjacent pens are going to look like. Start with a large rectangle with some length (L) and some width (W). You then have to break it up into five separate sections by drawing in four (4) more horizontal lines (in other words you have just added four more fences).
The perimeter formula now becomes
P = 2L + 6W instead of the regular P=2L+2W.
That should solve the problem.