Show work please. Find the dimensions of the rectangular region of area 200 in2

Question

Show work please.

Find the dimensions of the rectangular region of area 200 in2 with least perimeter. It costs $8 per linear foot of fencing. What is the maximum rectangular area that can be fenced for $4000? Assume the cost of constructing an oil transportation pipeline is $500,000 per kilometer on land and $1,800,000 per kilometer under water. Find the route that minimizes the cost of construction of a pipeline from a rig that is 10 km off-shore to a refinery that is 5 km down shore. Find the dimensions and the maximum volume of a open rectangular box that can be constructed from a 18" times 42" rectangular piece of cardboard by removing four squares at the four corners.

Explanation / Answer

1.

A = xy = 200 is fixed.

So y = 200/x

P = 2x + 2y = 2x+400/x

Minimize P. Differentiate with respect to x

dP/dx = 2 - 400/x^2 = 0

400/x^2 = 2

x^2 = 200

x = + sqrt(200) (ignore negative root, since x can't be negative).

Noting that d^2 P / dx^2 = 800/x^3 > 0 for all x>0, P is concave up, and thus our critical point is the location of minimum perimeter.

Since x = sqrt(200), we have y = 200/sqrt(200) = sqrt(200). So the rectangle is actually a square sqrt(200) by sqrt(200) inches. (about 14.14 by 14.14 inches)

2. A = xy. We want to maximize this.

But 8P = 4000

8(2x + 2y) = 4000

2x + 2y = 500

x + y = 250

y = 250 - x.   Plug this into A.

A = x(250-x) = 250x - x^2

Maximize A by finding critical points.

dA/dx = 250 - 2x = 0

2x = 250

x = 125

Note that the graph of A(x) is a downward opening parabola, so this is a maximum.

x+y = 150 implies that 125 + y = 250, or y = 125.

So the rectangle of maximum area that we can fence in for $4000 is a 125 ft by 125 ft square, of area 15625 sq ft.

3. If we land on shore 5-x km from the refinery, then the length of pipe is

(5-x) km on land + sqrt(10^2 + x^2) km underwater.

The cost of this is

C = 500,000 (5-x) + 1,800,000 sqrt(100 + x^2) = 2,500,000 - 500,000x + 1,800,000 sqrt(100 + x^2)

Take the derivative and set equal to 0

C' = -500,000 + 900,000 * [1/sqrt(100 + x^2)] * 2x = 0

Clear the denominator

-500,000 sqrt(100 + x^2)] + 1,800,000 x = 0

1,800,000 x = 500,000 sqrt(100 + x^2)

Square both sides

(1,800,000)^2 x^2 = (500,000)^2 * (100) + (500,000)^2 * x^2

x^2 = 8.3612

x = 2.89 km

So we should lay the pipe (5-x) = 5-2.89 = 2.11 km along the shore line on land, and then the rest of the pipe underwater to the rig.

4. V = l w h

If we cut x by x squares from the corners of the 18 x 42 piece of cardboard, then the height of the resulting box is x, the width is 42 - 2x, and the length is 18 - 2x

So V = x(42-2x)(18-2x) = x(756 - 120 x + 4x^2) = 756x - 120x^2 + 4x^3

Differentiate, set equal to 0

756 - 240x + 12 x^2 = 0

Quadratic formula yields

x = 3.917, 16.083. The latter value is too large (2x would then be greater than 18) So you can check that V is indeed maximized at x = 3.917 in (second derivative is -240 + 24x, whicn is negative for our value). The dimensions are 10.166 in by 34.166 in by 3.917 in, and the volume is 1360.50 cu in.

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