# Thread: having problems

1. ## having problems

I'm having some problems with calculus and this is really my last resort... if someone can help me its due in 3 hours...

A fence is to be built to enclose a rectangular area of 270 square feet. The fence along three sides is to be made of material that costs 6 dollars per foot, and the material for the fourth side costs 12 dollars per foot. Find the length and width (with ) of the enclosure that is most economical to construct.

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A parcel delivery service will deliver a package only if the length plus the girth (distance around) does not exceed 100 inches. Find the maximum volume of a rectangular box with square ends that satisfies the delivery company's requirements.

2. Originally Posted by abcdef
I'm having some problems with calculus and this is really my last resort... if someone can help me its due in 3 hours...

A fence is to be built to enclose a rectangular area of 270 square feet. The fence along three sides is to be made of material that costs 6 dollars per foot, and the material for the fourth side costs 12 dollars per foot. Find the length and width (with ) of the enclosure that is most economical to construct.

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i guess this answer is too late, but i just got on not so long ago.

we know that the area $LW = 270$, this is our constraint. call it equation (1)

we want to be as economical as possible, so let's make one of the shorter sides of the fence the \$12 per foot part.

thus, the cost for this side is 12W, and the cost for the other 3 sides is 6(2L + W), thus, the total cost C is,

$C = 12W + 6(2L + W) = 18W + 12L$

from equation (1), we have that $W = \frac {270}L$, thus

$C = 18 \cdot \frac {270}L + 12L = \frac {4860}L + 12L$ .......this is the function we want to minimize

thus we find C' as set it equal to zero and solve for L. then we can find W using the fact that W = 270/L

A parcel delivery service will deliver a package only if the length plus the girth (distance around) does not exceed 100 inches. Find the maximum volume of a rectangular box with square ends that satisfies the delivery company's requirements.
Let the length of the package be $y$ and the side length of the end be $x$

then the girth is $4x$ and the length is $y$

thus we want $4x + y \le 100$, this is our constraint

Now, the volume is given by $V = x^2 y$

from our constraint, we see that $y \le 100 - 4x$

thus, $V \le x^2 (100 - 4x)$ ........this is the function we want to maximize

thus we find $V'$ and set it equal to zero and solve for $x$, then we can find $y$

Note that, for the maximum, we use the = sign as opposed to the $\le$ sign

3. For some reason these problems i worked out and the answers r not working on web work... they are due in 50 minutes

4. Originally Posted by abcdef
For some reason these problems i worked out and the answers r not working on web work... they are due in 50 minutes
show me what you did and we'll see if we can fix it

5. k soo

LW=270
W=270/L

C=3240/L+12L
C1= -3240/L^2+12

L=+/- 16.43

which after that gave me a w of 1.. soo i don't no what i'm doing, my professor basically teaches us nothing

6. for the box problem

i got after taking the deriviative -12x^2+200x=0

-200 +/- sqrt(200^2-4(-12)(0)
divide by -24

x= 25/3

4(25/3)+y=100

y= 200/3

7. Originally Posted by abcdef
k soo

LW=270
W=270/L

C=3240/L+12L
C1= -3240/L^2+12

L=+/- 16.43

which after that gave me a w of 1.. soo i don't no what i'm doing, my professor basically teaches us nothing
shouldn't you leave the answer as exact? we have $W = L = \sqrt{270}$

8. after i kept it as the (sqrt(270)) it still says i'm incorrect

9. Originally Posted by abcdef
for the box problem

i got after taking the deriviative -12x^2+200x=0

-200 +/- sqrt(200^2-4(-12)(0)
divide by -24

x= 25/3

4(25/3)+y=100

y= 200/3
quadratic formula is not needed, and your answer is wrong

$200x - 12x^2 = 0$

$\Rightarrow 50x - 3x^2 = 0$

$\Rightarrow x (50 - 3x) = 0$

$\Rightarrow x = 0 \mbox{ or } 50 - 3x = 0$

now continue

Originally Posted by abcdef
after i kept it as the (sqrt(270)) it still says i'm incorrect
indeed, i made a mistake, my bad. i plugged in 12W as opposed to 18W, when we were substituting L. the actual equation for the cost is

$C = 18 \cdot \frac {270}L + 12L = \frac {4860}L + 12L$

now make the necessary changes

10. haha nice.. thank u soo much for ur lovely help