please help with math?

Question

Intro.
Charlie is a chicken farmer he wants to build a rectangular "free range" chicken pen with an Area of 150cm squared against an existing brick fence. He has 35m of fencing available to construct the OTHER three sides.

i. If the width of the pen is "x"m, find the length of the pen in terms of 'x'.

ii.SHow how you would find an expression for the area of the pen in terms of 'x'.

iii. Form an equation involving 'x' and the area of the pen.

iv. Show how you would solve this equation for 'x'

v.Find the POSSIBLE dimensions of the chicken pen,stating and restrictions on your answers.

vi.SHow other methods that may be used to calculate the dimensions of the pen;considering using technology, and again check that your values are appropriate.

vii. Which method do you prefer.Explain



can u please show work for all questions.. dont just write the answer.. thanks!!

Answer 1

Edited: With the new information.

Let x be the width and y be the length required.



Then we have the equations

xy = 150.................a

x + 2y = 35..............b

Then, y = (35-x)/2 (ans to i)



Substituting in equation a,

x(35-x)/2 = 150 (ans to ii and iii)

or x^2 -35x +300= 0

or (x-15)(x-20) = 0

Solving, x = 15 or 20 and correspondingly, y = 10 or 7.5 (iv)



this is the case when ALL of 35 meters of fencing is used. we can use LESS than 35 meters too.



Now, in this case, xy = 150

and 2y + x < 35

Substituting for y,

2 * 150 /x + x < 35

or x^2 - 35x +300 <0

or (x - 35/2) ^2 < 25/4............c



now if x > 35/2, inequation c becomes (on taking square roots)

x - 35/2 < 5/2

or x < 20



if x < 35/2 c becomes - (x - 35/2) < 5/2

or x > 15



Therefore the restriction is 15 < x < 20

and 7.5 < y < 10 (ans to v)

Answer 2

I think you have a little typo. What are the units for the area and perimeter? ill just assume you mean meters for both. So anyway,

Area = 150m

Perimeter = 35m

x= width and y = length



You can do another equation but this way is easier because you don't have to divide as much



The equation of perimeter is (x = width and y = length)

P = x + 2y (assuming the existing brick fence is the width)

P = 35 (given by fence length)



So 35 =x + 2y or



So y in terms of x is (35-x)/2 =y

and x in terms of y is 35 - 2y =x



Area = xy

so x =area/y --- x = 150/y



To solve for x use both equations for x and set them equal to each other

x = 35 - 2y = 150/y

and solve for y



What i did was graph both equations on my graphing calculator and found where they intersected. (where they intersect are the possible solutions). You could use the quadratic equation, but i was too lazy :)



So anyway my possible solutions were 7.5 =y (length)and x = 20 (width)

and also y = 10 and x = 15



One restriction is the fact that the dimensions have to be positive (not negative silly:) and the obvious restrictions are the set area and perimeter.

and

Answer 3

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Answer 4

according to the description you gave,we come to know that

(1)X*X'=150 so X'=150/X