QuestionA farmer has 150 feet of fence to enclose 1125 sq ft with squares of sides $x$ and $y$ . Find $x$ and $y$ . $x=$ $y=$

Studdy Solution

STEP 1

Assumptions1. The total length of the fence available is150 feet. . The total area to be enclosed is1125 square feet.
3. The region to be enclosed is in the shape of adjoining squares.
4. The big square has sides of length $x$ .
5. The small square has sides of length $y$ .

STEP 2

First, we need to set up the equations based on the given conditions. The total length of the fence is equal to the sum of the perimeters of the big square and the small square.
$4x +y =150$

STEP 3

The total area to be enclosed is equal to the sum of the areas of the big square and the small square.
$x^2 + y^2 =1125$

STEP 4

We have a system of two equations. We can solve this system by substitution or elimination. Let's solve it by substitution. First, solve the first equation for $y$ .
$y = \frac{150 -4x}{3}$

STEP 5

Substitute $y$ from the first equation into the second equation.
$x^2 + \left(\frac{150 -4x}{3}\right)^2 =1125$

STEP 6

implify the equation.
$x^2 + \frac{(150 -4x)^2}{9} =1125$

STEP 7

Multiply through by9 to clear the fraction.
$9x^2 + (150 -4x)^2 =10125$

STEP 8

Expand and simplify the equation.
$x^2 +22500 -1200x +16x^2 =10125$

STEP 9

Combine like terms.
$25x^2 -120x +22500 =10125$

STEP 10

Subtract10125 from both sides to set the equation equal to zero.
$25x^2 -1200x +12375 =0$

STEP 11

Divide through by25 to simplify the equation.
$x^ -48x +495 =0$

STEP 12

This is a quadratic equation in the form $ax^2 + bx + c =0$ . We can solve it by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula.
$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

STEP 13

Substitute $a =$ , $b = -48$ , and $c =495$ into the quadratic formula.
$x = \frac{48 \pm \sqrt{(-48)^2 -**495}}{2*}$

STEP 14

implify the equation.
$x = \frac{48 \pm \sqrt{2304 -1980}}{2}$

STEP 15

Calculate the value under the square root.
$x = \frac{48 \pm \sqrt{324}}{2}$

STEP 16

Take the square root.
$x = \frac{48 \pm18}{2}$

STEP 17

Calculate the two possible values for $x$ .
$x =33,15$

STEP 18

Substitute $x =33$ into the first equation to find the corresponding value for $y$ .
$y = \frac{150 -4*33}{3} =15$

STEP 19

Substitute $x =15$ into the first equation to find the corresponding value for $y$ .
$y = \frac{150 -4*15}{3} =30$ So, the solutions are $(x, y) = (33,15)$ and $(x, y) = (15,30)$ . The farmer can either make the big square with side length33 feet and the small square with side length15 feet, or the big square with side length15 feet and the small square with side length30 feet.

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