Algebra Help?

Question

I'm trying to help the girlfriend solve this problem however it has been many years since I have taken any college level math course. Unfortunately the book and online supplement associated with this lesson is very nonspecific. A solution and setup (explanation of how to find the solution) would be extremely helpful. Thank you!


Problem:

A rectangular field has a perimeter of 410 feet and is to have a n area at least 10,350 square feet. Within what bounds must the length, x, of the rectangular field lie? (hint: use the equation for perimeter in order to get a quadratic inequality for area in terms of the variable x only.)

Answer 1

Hi,

If the perimeter is 410, then one length and one width would add to 205. If the width is "x", then the length is "205 - x". The area of the rectangle would be found by x(205 - x). So the area, "y", is found by y = 205x - x². If I want to know when this has an area of an least 10,350 square feet, then let:

205x - x² = 10350

Factoring and solving this gives:

x² - 205x + 10350 = 0
(x - 115)(x - 90) = 0
x = 115 or x = 90

90 ≤ x ≤ 115

I hope this helps!! :-)

From 90 to 115 feet

Answer 2

P = 410 ft
Length = x ft
Width = (410/2 - x) ft = (205 - x) ft

A = x(205 - x)

10350 >/= 205x - x^2

x^2 - 205x + 10350 >/= 0

Let's first solve for
x^2 - 205x + 10350 = 0

So using the quadratic formula, we get
x = [205 +/- sqrt(42025 - 41400)]/2
x = [205 +/- sqrt(625)]/2
x= (205 +/- 25)/2

x = 115 or x = 90 (One is the length, the other the width)

To check, A = 115*90 = 10350 ft^2

Now you want the area to be AT LEAST this big. If the length and width are closer together, the area is bigger. If they are further apart, the area gets smaller, so
90 ft <= x <= 115 ft

I hope this is all clear!

Answer 3

A rectangular field will have two sides x and two sides y. 2x + 2y = 410. 2y = 410 - 2x. y = 205 - x

So we have x for one side and 205 - x for the other side.

x * (205-x) > 10350

205x - x^2 > 10350

x^2 - 205x + 10350 < 0

Let's solve to find out where x = 0

x = (205 +/- sqrt(205^2 - 4 * 1 * 10350)) / (2 * 1)

x = (205 +/- sqrt(42025 - 41400)) / 2

x = (205 +/- sqrt(625)) / 2

x = (205 +/- 25) / 2

x = 230/2 or x = 180/2

x = 115 or 90

The area will be greater than 10350 in the range (90, 115).

Answer 4

Area = width * length
perimeter = 2 (width + length)


they give both area and the perimter
10350 = w * L
410 = 2 (W + L)

solve for L
410 = 2 (W + L)
205 = W + L
L = 205 - W

plug the length back in
10350 = W * L
10350 = W (205 - W)
10350 = 205W - W^2
W^2 - 205W + 10350 = 0

use qudratic formula and you'll get W = 115 and 90

L = 205 - W
L = 205 - 115
L = 90

The width is 90ft
the length is 115ft

Answer 5

For increasing cubic binomials the final formulation is as follows: (a + b) ^ 3 = a^3 + 3*a^2*b^one million + 3*a^one million*b^2 + b^3 on your case, a is x and b is -y^5 So (x - y^5)^3 = x^3 + 3*x^2*(-y^5)^one million + 3*x^one million*(-y^5)^2 + (-y^5)^3 Simplified: =x^3 - 3x^2*y^5 + 3x*y^10 - y^15 :D

Answer 6

Perimeter:
2x+2y=410

Area:
x*y=10,350

2 equations and 2 unknowns

Answer 7

2X+2W = 410

X*W=10350

substituting,

2X^2 - 410X + 20700 = 0

using quadratic equation,

X range is (410 +/- 50)/4

or . . . 90 to 115 ft.

Answer 8

first of all say my girlfriend and get rid of the mullet