Calculus Problem?

A landscape architect plans to enclose a 2600 square foot rectangular region in a botanical garden. She will use shrubs costing $25 per foot along three sides and fencing costing $10 per foot along the north side.

(a) If x is the length of fencing used, write a formula for the total cost in terms of x only.

C(x) = ?

(b) Find the length of fence that will minimize the total cost

x= ?

(c) Find the minimum total cost. Give your answer to the nearest penny.

Calculus Problem?
Area = 2600 sq. ft. = (x)(y) -------- (1)

Perimeter = 2x + 2y ------------------(2)

Let x = length of fencing used, or one side of the rectangle.



Cost = $25 (2x + 2y - x) + $10 (x)

y = 2600/x from (1)

2x + 2y = 2x + 2(2600/x) = 2x + 5200/x



Cost = 25(2x +5200/x - x) + 10x

Cost = 50x + 130,000/x - 25x + 10x

(a) Cost (x) = 35x + 130,000/x



(b) dCost/dx = 0 = 35 + (-1)130,000 (x)^-2

130,000/x^2 = 35

x^2 = 130,000/35

x = +/- sqrt(130,000/35) = 60.94 feet %26lt;%26lt; answer



(c)

Minimum cost = 35(60.94) + 130,000/60.94

Minimum cost = $4,266.14 %26lt;%26lt; answer



Edit: To the answer above, you used the wrong equation for the cost:



" C(x)=25x+10y, eqn to minimize.

Use xy=2600 to solve for y: "



Here you are assigning x to be the sum of the 3 sides, and then you went on to get the area, 2600 = xy?? You are then multiplying the sum of the length of the 3 sides by the 4th side. That is NOT the area!
Reply:Area: A=xy=2600

C(x)=25x+10y, eqn to minimize.

Use xy=2600 to solve for y:

y=2600/x, now substitute into eqn.

C(x)=25x+10[(2600/y)]

Now do the differentiation tests to find minimum points. It should be straight forward.

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