# Thread: Help with some problems on Schaum Outline

1. ## Help with some problems on Schaum Outline

Ok so I'm reviewing algebra since it was the only math class I didn't do well in and thought I would work out all the problems in the Schaum's Outline. I have ran across 3 problems I'm not sure how to do.

a)Two numbers have the ratio 3:4. If 4 is added to each of the numbers the resluting ratio is 4:5. Find the two numbers.

b) Determine whether the graph of each relation is symmetric with respect to the y axis, x axis, or origin.

$\displaystyle y = (x-3)^3$

I want to work this one out rather than look at the graph.

$\displaystyle y = (x-3)^3$

$\displaystyle -y= (-x-3)^3$

$\displaystyle (-x-3)^3 = (-x)^3 -3(-x)^2(-3) +3(-x)(-3)^2 -(-3)^3$

$\displaystyle (-x-3)^3 = -x^3 + 9x^2 -27x +27$

$\displaystyle -y= -x^3 + 9x^2 -27x +27$

$\displaystyle y = x^3 - 9x^2 + 27x -27$

So it is symmetric with respect to the origin but is there a faster way of showing that. I tried going this route but got stuck

$\displaystyle -y = (-x-3)^3$

$\displaystyle -y = (-1)^3(x+3)^3$

and then got stuck.

c)Find the largest rectangle which can be inscribed in a right triangle whose legs are 6 and 8 inches repectively.

I know $\displaystyle A= xy$

and
$\displaystyle x^2 + y^2 = 100$

2. Originally Posted by 11rdc11
a)Two numbers have the ratio 3:4. If 4 is added to each of the numbers the resluting ratio is 4:5. Find the two numbers.
I would set this problem out as follows

$\displaystyle a:b = 3:4$

$\displaystyle a+4:b+4 = 4:5$

3. Originally Posted by pickslides
I would set this problem out as follows

$\displaystyle a:b = 3:4$

$\displaystyle a+4:b+4 = 4:5$
I tried that and didn't get the correct answer. Let me try again and see if I made an error.

4. Lol yep made a careless error. Thanks pickslides

5. I'm not sure what the answer book says but I arrived at $\displaystyle a=12$ and $\displaystyle b = 16$

6. Originally Posted by 11rdc11

c)Find the largest rectangle which can be inscribed in a right triangle whose legs are 6 and 8 inches repectively.

I know $\displaystyle A= xy$

and
$\displaystyle x^2 + y^2 = 100$

Largest rectangle implies find a maximum i.e $\displaystyle \frac{dA}{dx} = 0$

where $\displaystyle A= xy$ and $\displaystyle y = \sqrt{100-x^2}$

gives $\displaystyle A= x\times \sqrt{100-x^2}$ now find $\displaystyle \frac{dA}{dx} = 0$ and solve for x

7. Originally Posted by pickslides
Largest rectangle implies find a maximum i.e $\displaystyle \frac{dA}{dx} = 0$

where $\displaystyle A= xy$ and $\displaystyle y = \sqrt{100-x^2}$

gives $\displaystyle A= x\times \sqrt{100-x^2}$ now find $\displaystyle \frac{dA}{dx} = 0$ and solve for x
Ok I did exactly the same thing you just showed but I and ended up with $\displaystyle 5\sqrt{2}$

8. Originally Posted by 11rdc11
b) Determine whether the graph of each relation is symmetric with respect to the y axis, x axis, or origin.

$\displaystyle y = (x-3)^3$
The relation you posted is not symmetric to the origin.

$\displaystyle y = (x-3)^3$

$\displaystyle -y= (-x-3)^3$

$\displaystyle (-x-3)^3 = (-x)^3 -3(-x)^2(-3) +3(-x)(-3)^2 -(-3)^3$????
This is where you went wrong. For a lack of a better explanation, you "counted" the negatives twice. Look at the binomial cubic expansion in the general case:
$\displaystyle (a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$
$\displaystyle (a \;{\color{red}-}\;b)^3 = a^3 \;{\color{red}-}\; 3a^2 b + 3ab^2 \;{\color{red}-}\; b^3$

If you're saying that a = -x and b = -3, then you follow the first rule:
$\displaystyle (-x-3)^3 = (-x+({\color{red}-3}))^3 = (-x)^3 \;{\color{red}+}\;3(-x)^2(-3) +3(-x)(-3)^2 {\color{red}+}\;(-3)^3$

But, if you're saying that a = -x and b = 3, then you follow the second rule:
$\displaystyle (-x-3)^3 = (-x)^3 -3(-x)^2({\color{red}3}) +3(-x)({\color{red}3})^2 -({\color{red}3})^3$

You were combining both rules.

In either case,
$\displaystyle (-x-3)^3 = -x^3 - 9x^2 - 27x - 27$.
And since
$\displaystyle -y = -x^3 + 9x^2 - 27x + 27$,
the two polynomials are not equal, so this relation is not symmetric at the origin.

In fact, this relation is not symmetric to the x-axis, y-axis, or the origin. You'll now have to show algebraically that the first two are not the case.

01

9. Lol yes another careless error. I kept thinking this couldn't be symmetric with the origin but the back of the book said it was. Once again thanks for pointing out that I was using a negative 3 instead of a positive 3.

10. For the first problem.

$\displaystyle \frac{x}{y}=\frac{3}{4}$

$\displaystyle \frac{x+4}{y+4}=\frac{4}{5}$

Solving for y in the first gives

$\displaystyle \frac{4}{3}x=y$

Substituting into the second

$\displaystyle \frac{x+4}{\frac{4}{3}x+4}=\frac{4}{5}$

$\displaystyle x+4=\frac{16}{15}x+\frac{16}{5}$

$\displaystyle \frac{1}{15}x=\frac{4}{5}$

$\displaystyle x=12$

You can finish up.