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Math Help - Area Word Problems

  1. #1
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    Area Word Problems

    1) A rectangular garden is bounded on one side by a river and is to be fenced in on the three other sides. Fencing material for the side parallel to the river is $30 per foot. Materials for the two other sides cost $10 per foot. What are the dimensions of the garden of largest possible area if $1200 is to be spent for fencing material?

    2) A rectangular piece of cardboard measuring 20 inches by 32 inches is to be made into a box with an open top by cutting equal size squares from each corner and folding up the sides. Let x represent the length of a side of each square. What is the maximum volume of this box?
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  2. #2
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    1) may be not enough info.

    2) (20 - 2x)(32 - 2x)(x) cubic inches
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  3. #3
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    Quote Originally Posted by NRS11 View Post
    1) A rectangular garden is bounded on one side by a river and is to be fenced in on the three other sides. Fencing material for the side parallel to the river is $30 per foot. Materials for the two other sides cost $10 per foot. What are the dimensions of the garden of largest possible area if $1200 is to be spent for fencing material?

    2) A rectangular piece of cardboard measuring 20 inches by 32 inches is to be made into a box with an open top by cutting equal size squares from each corner and folding up the sides. Let x represent the length of a side of each square. What is the maximum volume of this box?
    1) let the distance of the fence opposite to the river be x and that of the adjacent side be y. Then 30x+20y=1200, isolate x, and the area is x*y. You only have to complete the square after i think
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  4. #4
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    Quote Originally Posted by NRS11 View Post
    1) A rectangular garden is bounded on one side by a river and is to be fenced in on the three other sides. Fencing material for the side parallel to the river is $30 per foot. Materials for the two other sides cost $10 per foot. What are the dimensions of the garden of largest possible area if $1200 is to be spent for fencing material?
    ...
    Let r be the distance of the fence parallel to the river.
    Let s be the length of the fence perpendicular to the river.
    Max Money: $1200 = $30*r + 2*$10*s
    Max Area: r*s


    1200 = 30r + 20s

    isolate r
     \dfrac{1200 - 20s}{30} = r

    substituting that in the area equation

     \text{Area = } \left ( \dfrac{ 1200 - 20s}{30} \right ) s

     \text{Area = } 40s - \dfrac{ 2s^2}{3}

    take the derivitive and determine the value of s when the slope is zero.

     0 = 40 - \dfrac{ 4s }{3}

    solve that to get the value of s.
    plug that back into the equation above to compute r.

    How that helps

    .
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