math help college math homework math 101

Question

you have 220 ft of fencing to enclose a rectangular plot that borders on a river. if you do not fence the side along the river, what is the largest area that can be enclosed?

Answer 1

This is ironically normally solved by Calculus.  However, this is also one of those sorts of problems that can be solved without Calculus if you have a good feelings for quadratic equations.

 

First, you have two unknowns here, which means to solve it completely, you are going to need two equations.  Your two unknowns are the length and width for the rectangular plot that borders the river.

 

There are two equations that should come naturally when dealing with geometry (whether rectangles, squares, circles, etc), and that is the formulas for the perimeters and areas.  You already know that you NEED the area in this problem, since that is what the problem asks for.  The area of a rectangle is given by: L*W = A , where L is the length and W is the width and A is the area.

 

The perimeter here is a little more tricky.  You are talking about the total FENCING being used.  One side of the fence does not need to be built since there is a river there! (Hoping our cattle don't swim away)  So instead of the usual perimeter equation of P = 2*L+2*W, we only care about the two widths and the one length for the fencing, so that P = L+2*W

 

You now have the two equations:

 

A = L*W

P = L+2*W = 220, since P = 220 ft

 

You can now substitute from the bottom equation for either L or W and plug into the top equation.  This should give you the formula for a parabola.  Since I don't think Math 101 is calculus, at this point you want to find the Vertex for the parabola.  The vertex will be the maximum point of the parabola this case and represent the maximum area possible.  You can also back out the width and length from the equations now.

Answer 2

A = xy

 

P = 2x + y = 220

 

y = 220 – 2x = 2(110 – x)

 

A = 2x(110 – x)

 

Zeros: x = 0 and x = 110

 

Axis of Symmetry: x = (110 – 0)/2 = 55

 

Vertex is on Axis of Symmetry and will be a maximum point because leading coefficient is negative so parabola is opening down.

 

Maximum Area = A(55) = 2*55(110–55)

= 2(50+5)(50+5)

 =2(2500+250+250+25)

= 5000+500+500+50 = 6050 ft^2

Answer 3

X - width

Y - Length

 

2X + Y = 220

 

 A =(220 - X ) X

 

 A = - X^2 + 220X  / quadratic with a<0 , the maximum value is at Vertex.

  

     Vertex( - 220/-4 , A ( -220/4) )

 

              ( 55 , 6050)

 

    Always maximum area that can be covered with a fence with length X unit is a square with side of X/4 length.