3 replies to this topic

[slightly]

Greetings. I need help from the ground up on this calculator. I have the fx-9750g PLUS. I figured out how to turn it on. I even graphed function. But, I have no idea what modes I need to complete my problems. I do not understand the terminology well enough to choose properly.

My problem is. a circle of radius 4 has a rectangle. the right upper vertex is on a point of the circle (x,y). x^2 + y^2 = 16 Express the Area of the rectangle as a function of x. I took the square root of 16-x^2=y to define y, A(x)=4x(sqrt 16-x^2). My problem is entering the function into the calculator. graphing it and finding the smallest positive value of x to maximize A. When I enter the function...it is not a rectangle. I have checked and that the formula is correct. Please help me learn to use the calculator.

Thanks Billie

[caspro]

the right upper vertex is on a point of the circle

Where are the other vertices ?

You can't work out anything if the only information you have is one vertex.
At the least, you need a definition of the diagonally opposite vertex.

Express the Area of the rectangle as a function of x.

This assumes that the rectangle changes, which assumes that you have a definition of the
rectangle, but you haven't given us a definition yet. All you've said is that one of it's points
is on a circle.

[slightly]

Thank you for replying.

I have determined the answer;however, I agree I did not represent the problem correctly. No excuse other than having been in a Pre-Cal book for 10 hours. This week marks the first time I have ever turned on a graphing calculator. For clarification:

the problem reads, a rectangle is inscribed in a circle of radius 4. (therefore each vertex is on the circle) the center of the circle is at the origin (this was not defined, however the examples show this to be assumed in this section) Let P= (x,y) be the point in quadrant I which is a vertex of R. the funtion is x^2+y^2=16 the goal here was to construct a function of x by redefining y. y^2=16-x^2 then substitute the new value of y to determine a Area as a funtion of x, perimeter as a function of x, and then enter those functions into a graphing calculator to determine the Max A with the smallest positive x.

I knew how to work the problem, just not my calculator. I have been a paramedic for 12 years and calculators were always off limits. the micrograms per drop per minute per kg of patient weight had to be figured in our heads due to the possibility of having a patient and no calculator. I do thank you all, I have much to learn about this machine and will ask many stupid questions in the future.

Chau,
bg

[caspro]

Ok, glad you've found your own answer.

One method of doing this would be to type:

FMax(4Xsqrt(16-X^2,0,4

giving the x and max values as:
[2.8284]
[31.999]

FMax is on OPTN' />F4' />F6' />F2' />