Math

[Thi]

hi people can you guys help me with this 2 math problems?

I'm learning about Applications of Quadratic functions , right now

We can determine the vertex of a quadratic function (y=ax^2 + bx + c) using (-b/2a, 4ac-b^2/4a)

Questions are:

1.What is the maximum rectangular area that can be enclosed with 256 m of fencing material? What are the dimensions of the rectangle?

2. George has decided to fence an area of the side of the garage so the dogs have someplace to play. If he has 12m of fencing and is going to use the garage as one of the sides of the pen, what is the maximum area of the rectangular area? What are the dimensions?

for each question i need 2 let statements

THANKS!!

[Trung Quốc]

1) let x be the lenght of the rectangular
and k is the width of the rectangular
we know that he have 256M of fencing material
so 2*(x+k)=256

Now solve for k in term of x

find the are of that rectangle= x*k , substitute the value you find above into k
then you will see that it is a quadratic equation. Now determine the max for that quadratic equation.

2) should be similar.

Have to go to class now. :)

QT

[Chris-T]

Since QT is busy now, I hope I can help you out a little (just in case you
did not get it already):

1)
Let x be the length of the rectangle and k is the width of the rectangle,
then ==> perimeter = 2*(x+k) = 256
==> k = 128 -x (eq1)
Let y be the Area ==> y = x*k = x*(128-x) ==>
y=-x^2 + 128x (eq2: your quad. eq w/ a=-1, b=128, and c=0).

The vertex of parabola is the max-point. Since c=0 ==>
vertex co-ordinates x= -b/2a = 64, y= -b^2/4a = 4096.
Answer: x= 64m, k=64m, and Area = 4096 m^2

2) Similar as (1) except the perimeter eq.
Let x be the length of the rectangle and k is the width of the rectangle,
then ==> perimeter = 2x + k = 12m (assumed the other side "k" is used
by the garage) ==> k = 12 -2x (eq1)
Let y be the Area ==> y = x*k = x*(12-2x) ==>
y=-2x^2 + 12x (eq2: your quad. eq w/ a=-2, b=12, and c=0).

The vertex of parabola is the max-point. Since c=0 ==>
vertex co-ordinates x= -b/2a = 3, y= -b^2/4a = 18.
Answer: x= 3m, k=6m, and Area = 18 m^2.

Double-check my math, plz. Hope you got the idea of finding Max of Y
(Y=quad. equation) using vertex formula.

[Thi]

thanks a lot guys!!!