Ved S. answered • 06/05/15

Math tutor for SAT, ACT, GMAT, GRE, middle and high school

2x + y = 8500 eq1

^{2}+ 8500x eq2

^{2}+ bx + c is at x=-b/2a

^{2}+ 8500(2125) = 9,031,250 meters

^{2}

Cassidy C.

asked • 06/05/15farmer ed has 8,500 meters of fencing, and wants to enclose a rectangular plot that borders on a river. if farmer ed does not fence the side along the river, what is the largest area that can be enclosed?

Follow
•
5

Add comment

More

Report

Ved S. answered • 06/05/15

Tutor

4.9
(9)
Math tutor for SAT, ACT, GMAT, GRE, middle and high school

Let's say the length of sides perpendicular to the river = x

and, the length of side parallel to the river = y

Since the side along the river is not fenced, only 3 sides remain to be fenced.

2x + y = 8500 eq1

2x + y = 8500 eq1

Area of the rectangle = x*y

substitute value of y from eq1

area = x(8500-2x)

=-2x^{2} + 8500x eq2

Since the coefficient of x2 is negative, this is a parabola opening down,

in which case it will have it's maximum value at the vertex.

Now, the vertex of parabola with standard form f(x) = ax^{2} + bx + c is at x=-b/2a

so, the vertex of parabola represented by eq2 will be at x = -8500/2(-2) = 2125

This means that area will be maximum when x = 2125

We plug this value in eq2 to get the maximum area.

Maximum area = -2(2125)^{2} + 8500(2125) = 9,031,250 meters^{2}

Tejas S. answered • 06/05/15

Tutor

4
(2)
Math, Physics & Chemistry Tutor - Alexandria, VA

2L + B = 8500

A=LB = L(8500-2L)= 8500L-2L^{2}

dA/dL = 8500 - 4L = 0

L = 2125 m

B = 8500 - 4250 = 4250 m

A=2125x4250=9031250 m^{2}

This is a classic optimization problem.

The formula for the area of a rectangle is A = L W (L= length, W = width)

We would like to optimize A, but A depends on two variables not just one.

To fix this problem, we use the equation 8500 = L + 2W. This is called the constraint equation. It comes about because the farmer must fence the width twice, but the length only once (the river is the other "fence").

The constraint equation can be reorganized as L = 8500 - 2W. Substituting this into the areas equation gives

A = (8500 - 2W) W = 8500 W - 2W^{2} .

This now depends on just one variable (W), so we can find the optimal value of W by differentiating this expression for A with respect to W and then setting the resulting derivative equal to zero.

A' = 8500 - 4W = 0 This implies that the optimal value of width is W = 8500 /4.

Substituting back into L = 8500 - 2W implies that the optimal value of length is L = 8500/2.

Notice that L turns out to be twice W.

The optimal area is just A = L W = (8500)^{2} /8

Ask a question for free

Get a free answer to a quick problem.

Most questions answered within 4 hours.

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.