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  1. #1
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    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.

    =

    =

    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.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by abcdef View Post
    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.

    =

    =
    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
    Last edited by Jhevon; November 3rd 2007 at 10:55 PM. Reason: slight error in the first question fixed :)
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  3. #3
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    For some reason these problems i worked out and the answers r not working on web work... they are due in 50 minutes
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by abcdef View Post
    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
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  5. #5
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    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
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  6. #6
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    for the box problem


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


    quadratic formula

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

    x= 25/3

    4(25/3)+y=100

    y= 200/3
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by abcdef View Post
    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}
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  8. #8
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    after i kept it as the (sqrt(270)) it still says i'm incorrect
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by abcdef View Post
    for the box problem


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


    quadratic formula

    -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

    Quote Originally Posted by abcdef View Post
    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
    Last edited by ThePerfectHacker; November 7th 2007 at 07:20 PM.
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  10. #10
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    haha nice.. thank u soo much for ur lovely help
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