# Practice problems help:

• Nov 13th 2013, 07:36 AM
cyrus106
Practice problems help:
#1
The top and bottom margins of a poster are 8 cm and the side margins are each 9 cm. If the area of printed material on the poster is fixed at 384 square centimeters, find the dimensions of the poster with the smallest area.

I did these steps:

xy = 384
y = 384/x

A = (x + 16) * (y + 18)
A = (x + 16) * (384/x + 18)

And I appear to be stuck at this step, the book doesn't have an examples listed for this problem either.

#2

Consider the function https://courses1.webwork.maa.org:808...910e13a9c1.png whose second derivative is https://courses1.webwork.maa.org:808...81ccadbc11.png. If https://courses1.webwork.maa.org:808...bdb7dea671.png and https://courses1.webwork.maa.org:808...9a52aab281.png, what is https://courses1.webwork.maa.org:808...910e13a9c1.png?

I've always had trouble with second derivatives, so an answer to this problem with an explanation would greatly help me! Much thanks again!
• Nov 13th 2013, 07:50 AM
SlipEternal
Re: Practice problems help:
#1 $\dfrac{384}{x} = 384x^{-1}$. You will eventually want to take a derivative, and it is much easier to avoid mistakes with the power rule than with the quotient rule.

$A(x) = (x+16)(384x^{-1}+18)$

Next, multiply it out. Do you remember the FOIL method? It tells you what to multiply. First Outside Inside Last (FOIL) means multiply the first terms: $x\cdot 384x^{-1}$, then the outside terms: $x\cdot 18$ then the inside terms: $16\cdot 384x^{-1}$ then the last terms: $16\cdot 18$. Add them all together. Then take the derivative and set it equal to zero. Next, take the second derivative and plug in any critical values you found. If the second derivative is positive at that value, you found a local minimum.

#2 I assume you are practicing integration. $\int f''(x)dx = f'(x) + C_1$ where $C_1$ is an arbitrary constant. Since you know $f'(0) = 3$, you can solve for $C_1$ after you integrate by plugging in $0$ for $x$. Then, you integrate again: $\int f'(x)dx = f(x) + C_2$. You are given $f(0)=4$, so you can solve for $C_2$ after this second integration by plugging in $0$ for $x$.

Note: If $\int f''(x)dx = f'(x)+C_1$ then $f'(x) = \int f''(x)dx - C_1$.