9. A fence is to be built around an area such as the one shown below, area a minimum and the perimeter is to be 300m? x x x y y 4y

Question

9. A fence is to be built around an area such as the one shown below. What must be the values of x and y if the area is to be a minimum and the perimeter is to be 300m?

inverted T-shape 

Bottom 4y

both ends y

stem of T X x X x X

Answer 1

Hi Helen,

     If I'm understanding the diagram correctly, you can set up the following 2 equations.

300 = 6y + 5x and A = 4y^2 + x^2. Proceeding as if this is correct, you can solve the first equation for either x or y. I'll solve for x.

x = (300-6y)/5 = 60- 6/5y. you can now substitute this into the area expression to get a function with only y's.

A(y) = 4y^2 + (60 -6/5y)^2.

 

Now, I don't know what class you're in so the method of solving it varies. This is a standard calculus problem so I'll do it that way. You could also simplify this and since you know this is a parabola, you can find the axis of symmetry, substitute this back in to find the minimum of the parabola, then plug in to find the other dimension.

 

So using calculus the derivative is A'(y) = 8y + 2(60 - 6/5y )* (-6/5). Simplifying you get A'(y) = 8y - 144 + 72/25y. Set it equal to 0 and solve. 272/25 y = 144 which if y = 225/17. Substituting this ugly number into x = 60- 6/5y you get x = 44and 2/17. I'm curious to check these numbers since they're so odd.

P= 6(225/17) + 5(750/17) = 1350/17 + 3750/17 = 5100/17 = 300 so that's right.

The area from my diagram is A = 225/17 * 900/17 + 750/17*750/17 = 202500/289 + 562500/289 = 765000/289 =2647 sq ft.

approximately.

If we round of y = 13, x = 44.4. These dimensions should produce a larger area since we claim we have the minimum area. So 

A = 13*52 + 44.4 *44.4 = 2647.36. I'm convinced. BUT I also think I'm not understanding your diagram.

 

Hope this helped in some way.

Answer 2

perimeter = 6y + 3X + 2x

4y = 3X

6y = (18/4)X

 

300 = (18/4)X + 3X + 2x = (30/4)X + 2x

 

300 - (30/4)X = 2x

 

x = 150 - (15/4)X

 

Area = 4y² + Xx = 4(3/4X)² + Xx = (9/4)X² + Xx

 

Area = (9/4)X² + X[(150 - (15/4)X]

 

Area = (9/4)X² + 150X - (15/4)X²

 

Area = -(6/4)X² + 150X

 

h = 50

k = 3750 sq meters is the maximum

There is no minimum

Answer 3

perimeter: 3x+6y+4y-x=300

2x+10y=300

y=30-.2x

area=4y^2+x^2

4(30-.2x)^2+x^2

4(900-12x+.04x^2)+x^2

3600-48x+.16x^2+x^2

3600-48x+1.16x^2

 

To minimize take the derivative and set it to zero

-48+2.32x=0

x=20.68

 

Use y=30-.2x=25.86

 

Hope that helps!

Deanna