Optimization Problem

[Jason76]

A farmer wants to fence an area of 6 million square feet in a rectangular field and then divide it in half with a fence parallel to one of the sides of the rectangle. What should the lengths of the sides of the rectangular field be so as to minimize the cost of the fence?

Given: Area

\(\displaystyle A = lw\)

\(\displaystyle 6,000,000 = lw\)

\(\displaystyle 6,000,000 = xy\)

Solve for y.

\(\displaystyle y = \dfrac{6,000,000}{x}\)

OR

\(\displaystyle y = (6,000,000)(x^{1})\)

Plug y into parameter equation

\(\displaystyle C = 3x + 2y\)

\(\displaystyle C = 3x + 2[(6,000,000)(x^{-1})]\)

\(\displaystyle C = 3x + 12,000,000 x^{-1}\) Is this right?

\(\displaystyle C' = \dfrac{d}{dx}[3x + 12,000,000 x^{-1}]\)

 

[DrPhil]

A farmer wants to fence an area of 6 million square feet in a rectangular field and then divide it in half with a fence parallel to one of the sides of the rectangle. What should the lengths of the sides of the rectangular field be so as to minimize the cost of the fence?

Given: Area

\(\displaystyle A = lw\)

\(\displaystyle 6,000,000 = lw\)

\(\displaystyle 6,000,000 = xy\)

Solve for y.

\(\displaystyle y = \dfrac{6,000,000}{x}\)

OR

\(\displaystyle y = (6,000,000)(x^{1})\)

Plug y into parameter equation

\(\displaystyle C = 3x + 2y\)

\(\displaystyle C = 3x + 2[(6,000,000)(x^{-1})]\)

\(\displaystyle C = 3x + 12,000,000 x^{-1}\) Is this right? YES

\(\displaystyle C' = \dfrac{d}{dx}[3x + 12,000,000 x^{-1}]\)

If you give x and y in units of 1000 ft, it will be a little easier to work with. Then \(\displaystyle xy = 6\) and

\(\displaystyle C = 3x + 2y = 3x + 12x^{-1}\)

Proceed with setting the derivative to 0.

 

[Jason76]

A farmer wants to fence an area of 6 million square feet in a rectangular field and then divide it in half with a fence parallel to one of the sides of the rectangle. What should the lengths of the sides of the rectangular field be so as to minimize the cost of the fence?

Given: Area

\(\displaystyle A = lw\)

\(\displaystyle 6,000,000 = lw\)

\(\displaystyle 6,000,000 = xy\)

Solve for y.

\(\displaystyle y = \dfrac{6,000,000}{x}\)

OR

\(\displaystyle y = (6,000,000)(x^{1})\)

Plug y into parameter equation

\(\displaystyle C = 3x + 2y\)

\(\displaystyle C = 3x + 2[(6,000,000)(x^{-1})]\)

\(\displaystyle C = 3x + 12,000,000 x^{-1}\) Is this right?

\(\displaystyle C' = \dfrac{d}{dx}[3x + 12,000,000 x^{-1}]\)

\(\displaystyle C' = 3 + (-12,000,000x^{-2})\) Continuing on

\(\displaystyle C' = 3 -12,000,000x^{-2})\)

\(\displaystyle 3 - 12,000,000x^{-2} = 0\)

\(\displaystyle -12,000,000x^{-2} = -3\)

\(\displaystyle x^{-2} = \dfrac{-3}{-12,000,000}\)

\(\displaystyle (x^{-2})^{-\dfrac{1}{2}} = (\dfrac{-3}{12,000,000})^{-\dfrac{1}{2}}\)

\(\displaystyle (x^{-2})^{-\dfrac{1}{2}} = (\dfrac{(-3)^{-\dfrac{1}{2}}}{(12,000,000)^{-\dfrac{1}{2}}})\)

\(\displaystyle x = (\dfrac{(-3)^{-\dfrac{1}{2}}}{(12,000,000)^{-\dfrac{1}{2}}})\) Have a feeling whatever this come out to, won't be correct on the computer.

 

[HallsofIvy]

A farmer wants to fence an area of 6 million square feet in a rectangular field and then divide it in half with a fence parallel to one of the sides of the rectangle. What should the lengths of the sides of the rectangular field be so as to minimize the cost of the fence?

Given: Area

\(\displaystyle A = lw\)

\(\displaystyle 6,000,000 = lw\)

\(\displaystyle 6,000,000 = xy\)

Solve for y.

\(\displaystyle y = \dfrac{6,000,000}{x}\)

OR

\(\displaystyle y = (6,000,000)(x^{1})\)

You mean \(\displaystyle (6,000,000)(x^{-1})\)

Plug y into parameter equation

\(\displaystyle C = 3x + 2y\)

\(\displaystyle C = 3x + 2[(6,000,000)(x^{-1})]\)

\(\displaystyle C = 3x + 12,000,000 x^{-1}\) Is this right?

\(\displaystyle C' = \dfrac{d}{dx}[3x + 12,000,000 x^{-1}]\)

\(\displaystyle C' = 3 + (-12,000,000x^{-2})\) Continuing on

\(\displaystyle C' = 3 -12,000,000x^{-2})\)

\(\displaystyle 3 - 12,000,000x^{-2} = 0\)

\(\displaystyle -12,000,000x^{-2} = -3\)

\(\displaystyle x^{-2} = \dfrac{-3}{-12,000,000}\)

\(\displaystyle (x^{-2})^{-\dfrac{1}{2}} = (\dfrac{-3}{12,000,000})^{-\dfrac{1}{2}}\)

The sign is incorrect. The two negatives in the numerator and denominator in the previous line cancel.

\(\displaystyle (x^{-2})^{-\dfrac{1}{2}} = (\dfrac{(-3)^{-\dfrac{1}{2}}}{(12,000,000)^{-\dfrac{1}{2}}})\)

\(\displaystyle x = (\dfrac{(-3)^{-\dfrac{1}{2}}}{(12,000,000)^{-\dfrac{1}{2}}})\) Have a feeling whatever this come out to, won't be correct on the computer.

Again, your sign is wrong. This is the square root of a positive number not negative. Also it will help a lot to recognise that 12,000,000= 3(4,000,000) and 4,000,000 is a perfect square.

 

[Deleted member 4993]

A farmer wants to fence an area of 6 million square feet in a rectangular field and then divide it in half with a fence parallel to one of the sides of the rectangle. What should the lengths of the sides of the rectangular field be so as to minimize the cost of the fence?

Jason, Thy name is "convoluter".....

L = length of the field in 1000 ft

W = width of the field in 1000 ft ..... assume the divider is parallel to W

A(rea) = L * W = 6 ........................................(1)

F(encing) = 2*L + 3* W.................................(2)

form (1) → W = 6/L .......................................(3)

using (3) in (2)

F = 2*L + 18/L

\(\displaystyle \frac{dF}{dL}\) = 2 - 18/L²

for minimum 'F' → \(\displaystyle \frac{dF}{dL} = 0 \) →

2 - 18/L² = 0 → 2 = 18/L² → L² = 9 → L = 3

Using (1) → W = 2

Length of the field = 3000 ft

Width of the field = 2000 ft

 

[Jason76]

A farmer wants to fence an area of 6 million square feet in a rectangular field and then divide it in half with a fence parallel to one of the sides of the rectangle. What should the lengths of the sides of the rectangular field be so as to minimize the cost of the fence?

Given: Area

\(\displaystyle A = lw\)

\(\displaystyle 6,000,000 = lw\)

\(\displaystyle 6,000,000 = xy\)

Solve for y.

\(\displaystyle y = \dfrac{6,000,000}{x}\)

OR

\(\displaystyle y = (6,000,000)(x^{1})\)

Plug y into parameter equation

\(\displaystyle C = 3x + 2y\)

\(\displaystyle C = 3x + 2[(6,000,000)(x^{-1})]\)

\(\displaystyle C = 3x + 12,000,000 x^{-1}\) Is this right?

\(\displaystyle C' = \dfrac{d}{dx}[3x + 12,000,000 x^{-1}]\)

\(\displaystyle C' = 3 + (-12,000,000x^{-2})\) :wink: Continuing on

\(\displaystyle C' = 3 -12,000,000x^{-2})\)

\(\displaystyle 3 - 12,000,000x^{-2} = 0\)

\(\displaystyle -12,000,000x^{-2} = -3\)

\(\displaystyle x^{-2} = \dfrac{-3}{-12,000,000}\)

\(\displaystyle x^{-2} = \dfrac{3}{12,000,000}\)

\(\displaystyle (x^{-2})^{-\dfrac{1}{2}} = (\dfrac{3}{12,000,000})^{-\dfrac{1}{2}}\) Fixed problem.

\(\displaystyle (x^{-2})^{-\dfrac{1}{2}} = (\dfrac{(3)^{-\dfrac{1}{2}}}{(12,000,000)^{-\dfrac{1}{2}}})\)

\(\displaystyle x = (\dfrac{(3)^{-\dfrac{1}{2}}}{(12,000,000)^{-\dfrac{1}{2}}})\)

\(\displaystyle x = \dfrac{(12,000,000)^{1/2}}{(3)^{1/2}}\) Final answer - unless it needs more simplification and/or rewritten as square roots.

\(\displaystyle x = \dfrac{\sqrt{12,000,000}}{\sqrt{3}}\)

 

[Jason76]

Thanks Khan, it was correct. I will go thru my calculations sometime to see if I can get the same.