# Math Help - quadratic function application

I came here about a month ago asking a question and you guys were very helpful. Now, I need help again. These questions are part of a project, but the questions deal with quad. functions/optimization:

1) You want to fence off a rectangular garden plot by putting a short brick wall along one edge and wooden fencing along the other three edges. The brick wall will cost $8 per linear foot, while the wooden fence costs$2 per linear foot.

a) Find a function that gives the total cost of the material in terms of the variable l and w; ignore the thickness of the brick.

b) If you have only $500 to spend on materials, what are the dimensions of the largest (biggest area) plot you can enclose? And the second: 2) A box is made from a 16 inch by 28 inch piece of cardboard by cutting equal squares from each corner and folding up the sides. a) Write a function expressing the volume of the box as a function of x. b) Determine the domain of this function c) Determine the size of the square that will produce a maximum volume for the box. Give a complete explanation of your solution. I understand it's a lot, but I would appreciate any sort of help. I'm really struggling with this. Thanks. - Paul 2. Originally Posted by paul I came here about a month ago asking a question and you guys were very helpful. Now, I need help again. These questions are part of a project, but the questions deal with quad. functions/optimization: 1) You want to fence off a rectangular garden plot by putting a short brick wall along one edge and wooden fencing along the other three edges. The brick wall will cost$8 per linear foot, while the wooden fence costs $2 per linear foot. a) Find a function that gives the total cost of the material in terms of the variable l and w; ignore the thickness of the brick. Let's say the brick wall is one of the "length" sides. The equation for perimeter is: P = l + l + w+ w And since the perimeter is also the amount of fence, the cost of the fence is: C =$2l + $8l +$2w + $2w b) If you have only$500 to spend on materials, what are the dimensions of the largest (biggest area) plot you can enclose?
Well then, you have the equation: 500 = $2l +$8l + $2w +$2w

Substitute: 500 = 2l + 8l + 2w + 2w

Add: 500 = 10l + 4w

Subtract 10l from both sides: 500 - 10l = 4w

Divide both sides by 4: 125 - 2.5l = w

Now you want to find the largest area, so the equation for Area is: A = lw

Substitute: A = l(125-2.5l)

So: A = 125l - 2.5l^2

Let's say that A = y and l = x: y = 125x - 2.5x^2

Now graph that and find what value of x gives the largest area.

3. Originally Posted by Quick

Now graph that and find what value of x gives the largest area.

If a>0
Then,
y=ax^2+bx+c
Is a parabola that opens up.
Thus it has a minimum value.
Which is located at,
x=-b/(2a)
(Actually that is where derivative is zero).

If a<0
Then,
y=ax^2+bx+c
Is a parabola that opens down.
Thus it has a maximum value.
Which is located at,
$$x=-b/(2a) Thus, to find a max/min for "a" non-zero You need to compute, -b/(2a) And determine what the sign of 'a' is. 4. Originally Posted by paul I came here about a month ago asking a question and you guys were very helpful. Now, I need help again. These questions are part of a project, but the questions deal with quad. functions/optimization: 1) You want to fence off a rectangular garden plot by putting a short brick wall along one edge and wooden fencing along the other three edges. The brick wall will cost 8 per linear foot, while the wooden fence costs 2 per linear foot. a) Find a function that gives the total cost of the material in terms of the variable l and w; ignore the thickness of the brick. b) If you have only 500 to spend on materials, what are the dimensions of the largest (biggest area) plot you can enclose? And the second: 2) A box is made from a 16 inch by 28 inch piece of cardboard by cutting equal squares from each corner and folding up the sides. a) Write a function expressing the volume of the box as a function of x. b) Determine the domain of this function c) Determine the size of the square that will produce a maximum volume for the box. Give a complete explanation of your solution. I understand it's a lot, but I would appreciate any sort of help. I'm really struggling with this. Thanks. - Paul Hello, Paul, you have to cut off 4 equal squares which have the side x. Thus the greatest square to cut of has a side length of 8". The base area of the box is: A=(16-2x)*(28-2x)=4x^2-88x+448, 0 ≤ x ≤ 8 the volume of the box is: V=x * A=4x^3-88x^2+448x The volume will become an extreme value (Minimum or maximum) if the first derivative equals zero: V'(x)=12x^2-176x+448. Now calculate: 0=12x^2-176x+448. Use the formula to solve a quadratic equation. I've got the results: x=22/3-2/3*sqrt(37) ≈ 3.278...$$
or
x=22/3+2/3*sqrt(37) ≈ 11.388...[/tex]

The 2nd value doesn't belong to the domain of the volume function.

Plug in the 1st value into the volume function and you'll get the greatest Volume: V = 663.851 cubic inches.

EB

PS.: I've got some troubles to use Latex, so I send you this text as plain text. I hope you can use it.

5. Hi,

I've attached a sketch of your problem.

EB

6. Originally Posted by earboth
Hi,

I've attached a sketch of your problem.

EB
Here's a good sketch of the problem