- Tue Feb 11, 2014 8:20 am
- Forum: Calculus
- Topic: The stiffness of a rectangular beam varies as its breadth...
- Replies:
**2** - Views:
**2281**

Thank you for putting me on the right track. Solving the derivative of: (144-h^2)^0.5 * h^3 = 0, this gives: 3 h^2(144-h^2)^0.5 - h^4(144-h^2)^-0.5 = 0 (by the poduct rule). Solving this eqn: h = -6*sqrt(3) and h = 6*sqrt(3), (only the positive value is valid). So the height should be 10.4 inches an...

- Sun Feb 09, 2014 10:04 am
- Forum: Calculus
- Topic: The stiffness of a rectangular beam varies as its breadth...
- Replies:
**2** - Views:
**2281**

Hi, can someone help me getting started with this problem?

The stiffness of a rectangular beam varies as its breadth and as the cube of its hight. Find the dimensions of the stiffest beam which can be cut from a circular log 12 inches in diameter.

Thanks.

Luke.

The stiffness of a rectangular beam varies as its breadth and as the cube of its hight. Find the dimensions of the stiffest beam which can be cut from a circular log 12 inches in diameter.

Thanks.

Luke.

- Fri Nov 29, 2013 9:07 am
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Finding the graph of an equation
- Replies:
**6** - Views:
**2903**

Thanks for your hint and showing me the solution on the Wolfram website.

Part of the implicit plot is shown there.

This plot is what I was after.

How can I get this same plot on my graphic calculator (if this is possible)?

Thanks for the answer.

Luke.

Part of the implicit plot is shown there.

This plot is what I was after.

How can I get this same plot on my graphic calculator (if this is possible)?

Thanks for the answer.

Luke.

- Thu Nov 28, 2013 7:50 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Finding the graph of an equation
- Replies:
**6** - Views:
**2903**

If I could graph this equation, I would be able to see what points (x,y) are on this graph, so then I could derive the slopes of the tangent lines at several of these points. One solution is (2,1) so this surely is one point on the graph and the slope of the tangent line here is -(11/10) (after impl...

- Thu Nov 28, 2013 2:57 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Finding the graph of an equation
- Replies:
**6** - Views:
**2903**

How can I graph the following equation?

x^3 * y^2 = x * y^3 + 6

I don't know how to isolate y (to get y as a function of x).

Thanks.

Luke.

x^3 * y^2 = x * y^3 + 6

I don't know how to isolate y (to get y as a function of x).

Thanks.

Luke.

- Fri Jun 07, 2013 2:14 pm
- Forum: Intermediate Algebra
- Topic: Exponential system of eqn's.
- Replies:
**4** - Views:
**2707**

OK, the log rules helped me out, I found the solutions now, thanks for your help.

Greetings.

Luke.

Greetings.

Luke.

- Thu Jun 06, 2013 1:53 pm
- Forum: Intermediate Algebra
- Topic: Exponential system of eqn's.
- Replies:
**4** - Views:
**2707**

Sorry, but I still can't solve these eqn's so please help. This is what I have till now: ln (4 * e^(.2/y))= ln e ^(1/y)) => (1.386 * .2)/y ln e = (1/y) ln e ( with ln e = 1) So (1.386 * .2)/y = 1/y . But these are clearly not equal to each other, so I must have done something wrong, but I can't figu...

- Tue Jun 04, 2013 5:48 pm
- Forum: Intermediate Algebra
- Topic: Exponential system of eqn's.
- Replies:
**4** - Views:
**2707**

How can I solve this system? (e = natural log)

x * e ^ -(0.2/y) = 4 (1)

x * e ^ -(1/y) = 1 (2)

I know that x= 5.65 and y= .577 but how do I get these values?

From the second eqn. I get:

ln x = 1/y, but how do I solve x = 4 * e ^(.2/y) ?

Thanks.

x * e ^ -(0.2/y) = 4 (1)

x * e ^ -(1/y) = 1 (2)

I know that x= 5.65 and y= .577 but how do I get these values?

From the second eqn. I get:

ln x = 1/y, but how do I solve x = 4 * e ^(.2/y) ?

Thanks.

- Tue Sep 18, 2012 7:14 pm
- Forum: Intermediate Algebra
- Topic: Solving a system of two eqn's.
- Replies:
**3** - Views:
**1891**

How to solve the following system of two equations?

(13/x) - (2/y) = - 0.3

(14.6/x) - (0.4/y) = 0.04

Thanks.

(13/x) - (2/y) = - 0.3

(14.6/x) - (0.4/y) = 0.04

Thanks.

- Wed Mar 14, 2012 6:46 pm
- Forum: Intermediate Algebra
- Topic: solve irrational eqn.: (x-9)^(1/3) - (x+26)^(1/3) + 5 = 0
- Replies:
**3** - Views:
**1621**

Hi, I checked the equation and there are no errors, when I solve this eqn. with my graphic calculator (TI89) it gives me two roots 1 and -18, but sadly this thing only gives me the results and not how to get to them. According to my textbook it should be done like you mentioned by cubing both member...