View Full Version : optimal dimensions with algebra and t1-83 calculator

LindaG

06-05-2011, 09:15 AM

Hi,

I am stuck on this question and trying to enter it in the t1-83 calculator.

Equation is 30L-L + 30W-W

I am not sure what my windows settings will be to graph this.

Please help :D

Denis

06-05-2011, 12:22 PM

Equation is 30L-L + 30W-W

That's NOT an equation; an equation has an equal sign.

What exactly are you asking? Do you realise that 30L - L = 29L and 30W - W = 29W?

Please post the original question (from your math book?) IN FULL.

LindaG

06-05-2011, 02:25 PM

A rectangular pen with a perimeter of 60ft. is needed to house three pigs.

a) Determine an equation to model the area of this pen

b)Using the T1-83 Plus, graph the relationship between the length of a side and the area of the rectangle. Create a window and fill in the missing window entries, then sketch the graph.

I thought i got the right equation for A) but now i am not so sure :S

LindaG

06-05-2011, 02:38 PM

A rectangular pen with a perimeter of 60ft. is needed to house three pigs.

a) Determine an equation to model the area of this pen.

b) Using the T1-83 Plus, graph the relationship between the length of a side and the area of the rectangle. Create a window and fill in the missing window entries, then sketch the graph.

mmm4444bot

06-06-2011, 10:13 AM

graph the relationship between the length of a side and the area of the rectangle

I think that they want a graph of the area function in terms of the rectangle's length. This means that you first need to express the width in terms of the length. In other words, you need an expression for W that contains the variable L.

To find this expression (i.e., the rectangle's width in terms of its length), write the perimeter equation, and then solve that equation for W.

Can you complete this first step?

The rectangle's area is expressed as L times W, so multiply your new expression for W by L, and you'll get a polynomial in L to graph.

Please show your work, from now on. 8-)

LindaG

06-08-2011, 03:49 PM

2x+2y=60

2y= 60-2x

2y=

y= 30-x

Area equation

A=LW

A=x(30-x)

A= 30x-x(squared)

This is what I have so far, i just can't figure it out :(

The windows settings on the t1-83 calculator is what i am stuck on...

mmm4444bot

06-08-2011, 08:38 PM

A= 30x-x(squared)

Very good. The area function A(x) is defined by the quadratic polynomial 30x - x^2

The windows settings on the t1-83 calculator is what i am stuck on...

On the TI-83, we're renaming L to x. And we're renaming A(x) to y, where x is the length of the rectangle and y is the corresponding area for the rectangle with that length.

You should know that the graph of y = -x^2 + 30x is a parabola that opens downward because (1) the graph of any quadratic polynomial is a parabola, and (2) the leading coefficient is negative (thus the parabola opens downward and the y-coordinate at the vertex point is a maximum).

Since we're interested only in that part of the graph which lies in Quadrant I (positive lengths and positive areas, right?), we can graph from one x-intercept to the other and set the maximum y value to include the parabola's vertex.

Solving the equation x(30 - x) = 0 gives us the intercepts:

(0, 0) and (30, 0)

The vertex point lies on the parabola's bisector (i.e., the vertical line passing halfway between the x-intercepts).

Halfway between 0 and 30 is 15.

Hence, the x-coordinate of the vertex is 15. The value of the expression x(30 - x) when x = 15 is the y-coordinate of the vertex point. Or, said another way, when one dimension of the pig pen is 15 units, the pen's area is 15(30 - 15) AND that area is the maximum.

Next, press the [WINDOW] button on your TI-83 and enter the following values :

Xmin = 0

Xmax = 30

Xscl = 1

Ymin = 0

Ymax = the value of x(30 - x) when x = 15

Yscl = 50

Press the [GRAPH] button.

8-)

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