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g100num [7]
3 years ago
11

A hardware store wants to enclose a 800 square foot rectangular area adjacent to its store to display gardening equipment. One s

ide will be formed by the wall of the store. The side opposite the store will be constructed with steel fencing costing $6 per foot and the other two sides will be constructed with wood fencing costing $3 per foot. What are the dimensions that will minimize cost
Mathematics
1 answer:
inna [77]3 years ago
8 0

Answer:

x = 20√2 ft and y = 40/√2 ft

Step-by-step explanation:

Let; x = length of steel fencing

y = length of a wood fence that is perpendicular to the store

Thus, since area is 800 ft², then;

xy = 800

Length of fence; L = x + 2y

From earlier, xy = 800

y = 800/x

Thus;

L = x + 2(800/x)

L = x + 1600/x

Now, steel fencing costing $6 per foot and the other two sides will be constructed with wood fencing costing $3 per foot. Thus, total cost is;

C(x) = 6x + 3(2y)

But y = 800/x. Thus;

C(x) = 6x + 3(1600/x))

C(x) = 6x + 4800/x

C'(x) = 6 - 4800/x²

At C'(x) = 0, the cost is minimized.

Thus=

6 - 4800/x² = 0

6x² = 4800

x² = 4800/6

x² = 800

x = √800

x = 20√2

When 0 < x < 20√2, C'(x) < 0, so we say that C(x) is decreasing

When x > 20√2, C'(x) > 0, so we say that C(x) is increasing

Thus, the cost is minimized when x = 20√2

Thus, putting 20√2 for x in y = 800/x, we have;

y = 800/(20√2)

y = 40/√2

Thus, dimensions that will minimize cost are;

x = 20√2 ft and y = 40/√2 ft

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