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26 October, 03:10

A regular garden of area of 75 ft^2 is bounded on three sides by a wall costing $8 per ft and on the fourth by a fence costing $4 per ft. what are the most economical dimensions of the garden?

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  1. 26 October, 05:43
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    It is a problem of maximums and minimums.

    Let's:

    x=large

    y=width

    xy=75 ⇒ y=75/x

    Lets;

    C (x, y) = cost of the fence and wall

    C (x, y) = 8 (x+2y) + 4x

    C (x, y) = 8x+16y+4x

    C (x, y) = 12x+16y

    C (x) = 12x+16 (75/x)

    C (x) = (12x²+1200) / x

    1) we have to do the first derivative:

    C' (x) = [x (24x) - (12x²+1200) ] / x²

    C' (x) = (24x²-12x²-1200) / x²

    C' (x) = (12x²-1200) / x²

    2) we get values when C' (x) = 0

    C' (x) = 0

    (12x²-1200) / x² = 0

    12x²-1200=0

    x=⁺₋√ (1200/12)

    x₁=10

    x₂=-10 (this solution is not valid).

    3) we have to do the second derivative:

    C'' (x) = 2400/x³

    C'' (10) = 2400 / (10³) >0 therefore, we have a minimun when x=10

    4) we find out the dimensions of the garden:

    x=10

    y=75/100=7.5

    Answer: the most economical dimensions of the garden would be:

    10 ft x 7.5 ft

    width=7.5 ft

    length=10 ft

    and the cost when the area is boundened would be: $240.

    C (x) = (12x²+1200) / x

    C (10) = $240
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