a few questions involving logs, limits, and minimums

[Guest]

I dont remember how to do this:

1) Solve the folowing: log₄(x + 2)+log₄(x - 3) = log₄(9)

2) Evaluate the limit if it exists:

. . .lim_x->a [(x + a)²/(x² + a²)]

The varible is throwing me off.

3) A cereal box in the shape of a rectangular prism is required to have a capacity of 5000 cm³. The thickness of the box must be 10 cm to allow for a comfortable grasp by most people. What dimensions of the box require the minimum amount of material? Ignore any overlap needed to join the faces of the box.

I know that the surface area is given by A = 2Lw + 2Lh + 2wh

The volume is V = Lwh

Is the "10 cm" the width?

V = 5000 cm³

How do you find the other two lengths?

Thanks for the help!

 

[skeeter]

\(\displaystyle log_4(x+2) + log_4(x-3) = log_4(9)\)
\(\displaystyle log_4[(x+2)(x-3)] = log_4(9)\)
\(\displaystyle (x+2)(x-3) = 9\)
can you take it from here? ... don't forget to check your solutions in the original equation.


limx-->a [(x+a)^2/(x^2+a^2)]
what's the problem? ... straight plug-in "a" for x
(a+a)^2/(a^2 + a^2) = 4a^2/2a^2 = 2


the cereal box problem ...

5000 = LWH , let W = "thickness" of 10 cm
so ...
5000 = L(10)H
500 = LH

A = 2LW + 2LH + 2WH
A = 20L + 2(500) + 20H
A = 20L + 1000 + 20H

since 500 = LH, H = 500/L ...

A = 20L + 1000 + 20(500/L) = 20L + 1000 + 10000/L

find dA/dL and minimize

 

[soroban]

Hello, bittersweet!

Here's #3 . . .

3) A cereal box in the shape of a rectangular prism is required to have a capacity of 5000 cm³.
The width of the box must be 10 cm to allow for a comfortable grasp by most people.
What dimensions of the box require the minimum amount of material?
(Ignore any overlap needed to join the faces of the box.)

                     L
              * - - - - - - *
            /             / |
        10/             /   |
        /             /     |
      * - - - - - - *       |
      |             |       | H
      |             |       |
      |             |       |
    H |             |       |
      |             |       *
      |             |     /
      |             |   /10
      |             | / 
      * - - - - - - *
             L

The volume is: \(\displaystyle V\:=\:L\,\times\,W\,\times\,H\)
So we have: \(\displaystyle \,L\,\times\,10\,\times\,H\:=\:5000\;\;\Rightarrow\;\;H\,=\,\frac{500}{L}\;\) [1]

Area of the front and back: \(\displaystyle \,2\,\times LH\;=\;2LH\)

Area of left and right: \(\displaystyle \,2\,\times 10H\;=\;20H\)

Area of top and bottom: \(\displaystyle \,2\,\times 10L\;=\;20L\)

\(\displaystyle \;\;\)Total area: \(\displaystyle \,A\;=\;20L\,+\,20H\,+\,2LH\)

Substitute [1]: \(\displaystyle \,A\;=\;20:\,+\,29\left(\frac{500}{L}\right)\,+\,2L\left(\frac{500}{L}\right)\)

\(\displaystyle \;\;\)and we have: \(\displaystyle \,A\;=\;20L\,+\,10,000L^{-1}\,+\,1000\)


Differentiate and equate to zero: \(\displaystyle \,A'\;=\;20\,-\,10,000L^{-2}\;=\;0\)

Multiply by \(\displaystyle L^2:\;\;20L^2\,-\,10,000\:=\:0\;\;\Rightarrow\;\;L^2\:=\:500\;\;\Rightarrow\;\;L\,=\,10\sqrt{5}\)

From [1]: \(\displaystyle \,H\:=\:\frac{500}{10\sqrt{5}}\:=\:10\sqrt{5}\)


Therefore: \(\displaystyle \,L\,=\,10\sqrt{5},\;\;W\,=\,10 ,\;\;H\,=\,10\sqrt{5}\)