Need help with 2 questions

• Dec 13th 2009, 05:14 PM
11rdc11
Need help with 2 questions
Hi guys for some reason I'm having problems with these 2 questions, Any help would be great.

1. For the equation $60x^3-3x^5 = y$ where does the largest slope occur?

Ok here is how I thought I would find the answer for this problem. If I take the second derivative and equal it to 0 and find my critical points, then one of my points will be the the largest slope. Is this correct?

$f(x) = 60x^3-3x^5$

$f'(x) = 180x^2 -15x^4$

$f''(x) = 360x -60x^3$

so

$0 = 60x(6-x^2)$

so

$x = 0, \sqrt{6}, -\sqrt{6}$

so I think my largest slope occurs at

$\sqrt{6}, -\sqrt{6}$

Is this correct?

2.A rectangular storage container with a top is to have a volume of $6 M^3$. The length of the base is 1.5 times the width. Material for the base costs $3 per square meter, material for the sides cost$5 per square meter and the material for the top costs \$8 per square meter. Find the dimensions of the container with minimal total cost.

I'm not sure how to do the second one.

I know

w=x
l=1.5

but what do I use for height?

Also I know I have to use surface area but can't come up with an equation without the height.
• Dec 13th 2009, 05:41 PM
11rdc11
Ok so I tried this approach but not sure if it right

w=x
l = 1.5x
h=y

so for volume

$6 = (x)(1.5x)(y)$

Then I solve for y

$y = 4x^2$

$SA = 2(1.5x)(x) +2(1.5x)(y) + 2(x)(y)$

Then sub y in

$SA= 2(1.5x^2) +2(1.5x)(4x^2) +2(x)(4x^2)$

$3x^2 +20x^3$

Then I'm lost after that lol
• Dec 13th 2009, 05:56 PM
Jhevon
Quote:

Originally Posted by 11rdc11
Ok so I tried this approach but not sure if it right

w=x
l = 1.5x
h=y

so for volume

$6 = (x)(1.5x)(y)$

Then I solve for y

$y = 4x^2$

$SA = 2(1.5x)(x) +2(1.5x)(y) + 2(x)(y)$

Then sub y in

$SA= 2(1.5x^2) +2(1.5x)(4x^2) +2(x)(4x^2)$

$3x^2 +20x^3$

Then I'm lost after that lol

should it not be $y = \frac 4{x^2}$?

and remember, you want the cost equation, not the surface area equation to be your objective

after you get that, just find the minimum points of that function
• Dec 13th 2009, 06:10 PM
11rdc11
Quote:

Originally Posted by Jhevon
should it not be $y = \frac 4{x^2}$?

and remember, you want the cost equation, not the surface area equation to be your objective

after you get that, just find the minimum points of that function

lol yep it should be

$y =\frac{4}{x^2}$

So the derivative of the SA is the Cost function? If not how do I set up the cost function?
• Dec 13th 2009, 06:12 PM
Jhevon
Quote:

Originally Posted by 11rdc11
lol yep it should be

$y =\frac{4}{x^2}$

So the derivative of the SA is the Cost function? If not how do I set up the cost function?

the cost for each "side" (can mean top or bottom) of the box is (cost of the side)*(area of the side)

think you can find the equation now?
• Dec 13th 2009, 06:19 PM
Jhevon
Quote:

Originally Posted by 11rdc11
Hi guys for some reason I'm having problems with these 2 questions, Any help would be great.

1. For the equation $60x^3-3x^5 = y$ where does the largest slope occur?

Ok here is how I thought I would find the answer for this problem. If I take the second derivative and equal it to 0 and find my critical points, then one of my points will be the the largest slope. Is this correct?

$f(x) = 60x^3-3x^5$

$f'(x) = 180x^2 -15x^4$

$f''(x) = 360x -60x^3$

so

$0 = 60x(6-x^2)$

so

$x = 0, \sqrt{6}, -\sqrt{6}$

so I think my largest slope occurs at

$\sqrt{6}, -\sqrt{6}$

Is this correct?

by the way, this was correct. these are the x-values where you have the maximum slope.
• Dec 13th 2009, 06:31 PM
11rdc11
Quote:

Originally Posted by Jhevon
by the way, this was correct. these are the x-values where you have the maximum slope.

Thanks so the x values of $\sqrt{6}, - \sqrt{6},$ but not the x value of 0 right?

The cost function would be

$C = 4.5x^2 +12x^2 +\frac{60}{x} + \frac{40}{x}$

correct?

Thanks for the help
• Dec 13th 2009, 06:43 PM
Jhevon
Quote:

Originally Posted by 11rdc11
Thanks so the x values of $\sqrt{6}, - \sqrt{6},$ but not the x value of 0 right?

that's right. x = 0 is a local minimum for the derivative function.

Quote:

The cost function would be

$C = 4.5x^2 +12x^2 +\frac{60}{x} + \frac{40}{x}$

correct?

Thanks for the help
yes, that's correct. you can write it as $C = 16.5x^2 + \frac {100}x$ though

now find the minimum points of that function. you will be able to use that to find the dimensions
• Dec 13th 2009, 06:45 PM
11rdc11
Quote:

Originally Posted by Jhevon
that's right. x = 0 is a local minimum for the derivative function.

yes, that's correct. you can write it as $C = 16.5x^2 + \frac {100}x$ though

now find the minimum points of that function. you will be able to use that to find the dimensions

Thanks that was my next step.