1. ## Word/equation question

I have done questions like this before, But i am not sure how to start this one. Once I have the beginning equation I should be able to do it. As for the second one we havent done much on this in class so I'm not sure where to start.
Any help would be great. Thanks

2. Originally Posted by stacyg
I have done questions like this before, But i am not sure how to start this one. Once I have the beginning equation I should be able to do it. As for the second one we havent done much on this in class so I'm not sure where to start.
Any help would be great. Thanks
1. If the total area is a = 3000 cm² and the width of the poster is x then you know:

$l = \dfrac{3000}x\ ,\ w = x$

2. Cut of the margin:
from the length cut of 12 cm in total and from the width cut of 4 cm in total. The area of the printing part is then:

$a_p = \left(\dfrac{3000}x - 12 \right) (x - 4) = -12x - \dfrac{12000}x + 3048$

3. Now calculate the maximum of $a_p$ using derivation.

--------------------------------

You are supposed to know that $speed = \dfrac{distance}{time}~\implies~time = \dfrac{distance}{speed}$

Therefore: Traveling time is $t = \dfrac{60}v$

Total costs: $C(v)= \underbrace{\left(\dfrac{60}v \right)}_{time}\cdot \overbrace{\left(\dfrac12 \cdot v^3 + 20 \underbrace{+ 3000}_{fix \ costs} \right)}^{costs\ per\ hour}$

Now calculate the maximum of $C(v)$ using derivation.

3. Thanks. I'm not sure if I've done the differentiation right but for the 1st question would x=12 therefore the area be 1904cm squared.

And for the second question would c(v)=150.

Thanks

4. Originally Posted by stacyg
Thanks. I'm not sure if I've done the differentiation right but for the 1st question would x=12 therefore the area be 1904cm squared.

And for the second question would c(v)=150.

Thanks
1. If the width of the "poster" is 12 cm then the length must be 250 cm, that means 2 and a half meter. That's a quite unusual format for a poster.
2. With C(v) = 150 you can pay 3 minutes of the fixed costs!

3. As you suspected there must be gone something wrong. Therefore I'll give you the equations which you must solve for x or v respectively:

$\dfrac{12000}{x^2} - 12 =0$

$60v-\dfrac{181200}{v^2} = 0$

Good luck!

5. Originally Posted by earboth
1. If the width of the "poster" is 12 cm then the length must be 250 cm, that means 2 and a half meter. That's a quite unusual format for a poster.
2. With C(v) = 150 you can pay 3 minutes of the fixed costs!

3. As you suspected there must be gone something wrong. Therefore I'll give you the equations which you must solve for x or v respectively:

$\dfrac{12000}{x^2} - 12 =0$

$60v-\dfrac{181200}{v^2} = 0$

Good luck!
So I solved for x and got 31.62 for the 1st question.
And for the second one I have no idea how to solve it.
My math skills arent the best but I need this paper for my civil engineering diploma. This is a great help.

6. Originally Posted by stacyg
So I solved for x and got 31.62 for the 1st question. <<<< OK
And for the second one I have no idea how to solve it.
My math skills arent the best but I need this paper for my civil engineering diploma. This is a great help.
I'll show you how you could solve the 2nd equation:

$60v-\dfrac{181200}{v^2} = 0$ Multiply through by v²:

$60v^3-181200 = 0$ Move the constant to the RHS by adding 181200

$60v^3 = 181200$ Divide through by 60:

$v^3 = 3020$ Now take the cube root on both sides:

$v = \sqrt[3]{3020} \approx 14.454$

You now know the optimal speed of the ferry. Plug in this value into C(v) to get the optimal costs.