1. ## Dimensions of Kindle (Electronic pad with a screen)

I created this problem and wanted to check my answers with the MathForum.

x = length
y = width

Scenario:
There are two Kindles.
K(1) has the dimensions of 3.6 in by 4.8 in. with a surface area (SA) of $17.28in^2$ and a diagonal that measures 6 in
K(2) has the dimensions of 5.4 in by 7.9 in. with a surface area of $42.66in^2$ and a diagonal that measures 9.56 in

a. How much bigger is K(2)'s SA compared to K(1)'s SA?

b. Write an equation that represents the maximum and minimum value of
x and y when the SA of K(3) is 42.66 in.
my answer: Either $x=42.66*y$ or $y=42.66*x$

Note: For 'c' I want to know if it's possible to have a minimum and maximum value of the length and width when the diagonal is 6 in. My guess is that you can only have 1 combination of 'x' and 'y' that gives you a SA of 42.66 and a diagonal of 6 in.

c. Write an equation/system of equations that represents the maximum and minimum value of x and y if the diagonal of K(3) is 6 inches.

my answer: $x^2 + y^2 = (6)^2$ --> y = -X + 6 (If I simplified that right) and $x=42.66*y$ or $y=42.66*x$; where the two lines intersect is the value of x and y when the diagonal is 6.

2. Originally Posted by Masterthief1324
I created this problem and wanted to check my answers with the MathForum.

x = length
y = width

Scenario:
There are two Kindles.
K(1) has the dimensions of 3.6 in by 4.8 in. with a surface area (SA) of $17.28in^2$ and a diagonal that measures 6 in
K(2) has the dimensions of 5.4 in by 7.9 in. with a surface area of $42.66in^2$ and a diagonal that measures 9.56 in

a. How much bigger is K(2)'s SA compared to K(1)'s SA?
Yes, that is correct.

b. Write an equation that represents the maximum and minimum value of
x and y when the SA of K(3) is 42.66 in.
my answer: Either $x=42.66*y$ or $y=42.66*x$
What? You said there were two kinds of "kindle", K(1) and K(2). Where did K(3) come from? Without any information about K(3) it is impossible to answer this. Did you leave something out?

Note: For 'c' I want to know if it's possible to have a minimum and maximum value of the length and width when the diagonal is 6 in. My guess is that you can only have 1 combination of 'x' and 'y' that gives you a SA of 42.66 and a diagonal of 6 in.

c. Write an equation/system of equations that represents the maximum and minimum value of x and y if the diagonal of K(3) is 6 inches.

my answer: $x^2 + y^2 = (6)^2$ --> y = -X + 6 (If I simplified that right)
No, you didn't. " $x^2+ y^2= z^2$" does NOT reduce to "x+ y= z". For example, $4^2+ 3^2= 16+ 9= 25= 5^2$ but $4+ 3\ne 5$.

and $x=42.66*y$ or $y=42.66*x$; where the two lines intersect is the value of x and y when the diagonal is 6.
If $x^2+ y^2= 36$ and xy= 42.66, then y= 42.66/x so $x^2+ \frac{1819.8756}{x^2}= 36$. Multiplying both sides by $x^2$, $x^4+ 1819.8756= 36x^2$ or $x^4- 36x^2+ 1819.8756= 0$. If you let $u= x^2$, this becomes the quadratic equation $u^2- 36u+ 1819.8756= 0$. Solve that for u, then take the square root to find x.

But are you sure you have interpreted the question correctly? The problem said "find maximum and minimum" values, not specific values.

3. Sorry, I did make that last problem confusing.
I hope this clarifies it.

K(3) is a hypothetical Kindle device with a surface area of $42.66in^2$ and a diagonal of 6 in.

c1. Can you have more than one dimensions that satisfy the SA & the diagonal length of 6in? Why or why not?
my answer: Not if the diagonal is 6 inch -- there is only 1 value of x and y (length and width)

c2. Can you represent c1. with and equation?
my answer: Since the diagonal is 6, if I was to use the Pythagorean theorem, there is only 1 value of x and y that would fit the specifications.

(specifications)
You know indefinitely that the K(3) has a surface area of 42.66 and a diagonal of 6 in:

my equation:
x^2 + y^2 = 6 (is it 6 or 6^2?) --> y=√(z^2 - x^2)
x*y = 42.66

If I was to graph the two equations, will where they intersect be the dimensions that fit the specifications?

4. b. Write an equation that represents the maximum and minimum value of
x and y when the SA of K(3) is 42.66 in.
If K(3) has a surface area of 42.66, then you know that

$xy=42.66$

since the diagonal is 6 inches, you know that

$x^2 + y^2 = 6^2$ (draw a picture to understand why it is 6^2 and not 6)

You have two equations, you can try to solve them both for y and graph them to approximate a solution

$y=42.66/x$
$y=sqrt(36-x^2)$

Once you look at this graph you will realize that there is no solution to your problem

K(3) is physically impossible

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To understand this, think of the kindle with diagonal 6 that has the largest surface area, this kindle would be square. It would have x = y = sqrt (18), so its largest surface area would be xy=18

I created this problem and wanted to check my answers with the MathForum.
To fix your problem, you must either increase the size of the diagonal or reduce the surface area of K(3) to make the two curves intersect