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

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

18.75 gallons

step-by-step explanation: