Second Part Final (4 Problems Need Help)

**OverclockerR520** · May 16th 2006, 09:59 PM

Hello,

Well I think I did fairly well on my first part of the math final. Well I've been studying for my second part of the final which is wednesday night. I have 4 problems this time I need help with, everything else is going smooth. Thanks for the help on the previous post i understand those pics were hard to make out lol. Well anyway here goes nothing...

#28) A fourth grade class decides to enclose a rectangular garden using the side of the school as one side of the rectangle. What is the maximuum area that the class can enclose with 32 feet of fence? What should be the dimensions of the garden be in order to ield this area?

#30) Maximizing Area: The sum of the base and the height of a parallelogram is 69cm. Find the dimensions for which the area is a maximum.

#32) (confused me like crazy) Finding the Height of a Cliff: A water balloon is dropped from a cliff. Exactly 3 sec later, the sound of the balloon hitting the ground reaches the top of the cliff. How high is the cliff? (maybe this will help: t1 + t2 = 2) (maybe this will help too s=16t /\ 2 t is seconds and time is t1 that it takes the tablet to hit the water can be found as follows s=16(t1)/\2

lastly #38) Norman Window. A Norman window is a rectangle with a semicircle on top. Sky blue windows is designing a Norman window that will require 24 feet of trim. What dimensions will allow the maximum amount of light to enter a house? (maybe this will help) when looking at a normal window you see a rectangle with a half circle attached to the top. the x-axis is the top of that rectangle flowing horizontally and the y-axis is the side of that rectangle flowing vertically.

any help would be much appreciated before 3:30pm wednesday eastern time. Sorry for the late notice its been a hectic week

**earboth** · May 17th 2006, 04:53 AM

Originally Posted by OverclockerR520

Hello,... Well anyway here goes nothing...

#28) A fourth grade class decides to enclose a rectangular garden using the side of the school as one side of the rectangle. What is the maximuum area that the class can enclose with 32 feet of fence? What should be the dimensions of the garden be in order to ield this area?...
Sorry for the late notice its been a hectic week

Hello,

to speed up with your hectic week:
length of rectangle : l
width of rectangle : w
area of rectangle : w*l

Part of the perimeter which consists of wire = 2w +l = 32'
That means l = 32'-2w

You get the area with respect to w:
$A(w)=(32-2w) \cdot w=-2w^2+32w$

Let the first derivative be zero and you'll get: w = 8', thus l = 16'

Maximal area = 128 sqrft.

Greetings

EB

**earboth** · May 17th 2006, 05:20 AM

Originally Posted by OverclockerR520

Hello,...

#32) (confused me like crazy) Finding the Height of a Cliff: A water balloon is dropped from a cliff. Exactly 3 sec later, the sound of the balloon hitting the ground reaches the top of the cliff. How high is the cliff? (maybe this will help: t1 + t2 = 2) (maybe this will help too s=16t /\ 2 t is seconds and time is t1 that it takes the tablet to hit the water can be found as follows s=16(t1)/\2 ...
Sorry for the late notice its been a hectic week

Hello,

you've got 2 movements:
an accelerated movement when the balloon is falling down. (I don't deal with air-resistance etc.)
a linear movement of the sound going up.

Let c be the height of the cliff. then you get:
$c={1 \over 2} \cdot 9.81 \cdot t_1^2$
$c=330 \cdot t_2$ . I presume it's a cold day with a temperature of 0°C. Then the velocity of sound through air is 330 m/s.

From both equations you can calculate the times which add up to exactly 3 seconds:
${c \over 330}+\sqrt{{2c \over 9.81}}=3$

After a few transformations ( isolate the square-root, square both sides of the equation: Be aware that you've got a binomial formula on the RHS!, solve the quadratice equation for c, take only the positive value!) you'll get:

c = 40.5986 m

Now I've to run: Catch some €s.

Greetings

EB

**ThePerfectHacker** · May 17th 2006, 05:41 AM

Originally Posted by OverclockerR520

Hello,

#30) Maximizing Area: The sum of the base and the height of a parallelogram is 69cm. Find the dimensions for which the area is a maximum.

You have $b+h=69$ with $b,h>0$ you need to maximize the function $A=bh$ Note that $h=69-b$
Thus,
$A=b(69-b)=69b-b^2$
This is a parabola, with a maximum point because the coefficeint in front of $b$ is negative. Find its turing point,
$-\frac{69}{-2}=39.5\mbox{cm}$
Thus,
$h=39.5\mbox{cm}$

**earboth** · May 17th 2006, 10:59 AM

Originally Posted by ThePerfectHacker

You have $b+h=69$ with $b,h>0$ you need to maximize the function $A=bh$ Note that $h=69-b$
Thus,
$A=b(69-b)=69b-b^2$
This is a parabola, with a maximum point because the coefficeint in front of $b$ is negative. Find its turing point,
$-\frac{69}{-2}=39.5\mbox{cm}$
Thus,
$h=39.5\mbox{cm}$

Hello,

have a look here:

I believe that you made a minor mistake:

$-\frac{69}{-2}=39.5\mbox{cm}\ \ \leftarrow \mbox{are you sure?}$

Greetings

EB

**topsquark** · May 17th 2006, 12:47 PM

Originally Posted by earboth

Let c be the height of the cliff. then you get:
$c={1 \over 2} \cdot 9.81 \cdot t_1^2$
$c=330 \cdot t_2$ . I presume it's a cold day with a temperature of 0°C. Then the velocity of sound through air is 330 m/s.

Do the same process, except replace the second line with:
$c={1 \over 2} \cdot 32 \cdot t_1^2$
and the third with:
$c=1100 \cdot t_2$

This will give you the answer in terms of feet, not meters.

-Dan

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