# Math Help - Optimization Homework Help

1. ## Optimization Homework Help

1. A sm long trough is in the shape of an isosceles triangular prism with dimensions 30cm for both the sides of the triangle's face and the length is 300cm. I need to find the width of the triangular prism that will maximize the volume of thr trough. Please show all steps. Recall: V=(A x-section) L

2. Two buildings, A and B, in a school need to be connected with a fibre optic cable. Building A is 70m (up/north) from a roadway and B is 200m down/along the roadway. The cable must be laid underground across the playing field, but along the roadway it can ve constructed above ground. If the cost of it underground is $1000/m while above ground it is$500/m, where should point C be located to minimize the total cost of laying the cable? (ADC is a right angle triangle, and C is a striaght line from B)

3. A future shop store sells on average 20 ipods per week at a price of $200 per ipod. In any given week, the store will have a maximum of 40 ipods for sale. Market research shows that for each$5 increase in price there will be 2 less sales per week, but also for each $5 decrease there will be 2 more sales per week. The cost of the ipods for the store is$100 per ipod. What should the sale price of each ipod be to maximize profit? What is the maximum profit? recall: p(x)= r(x)-c(x)

4. a corridor is uniformly 2m wide and makes a right angle. (opposite and adjacent sides of the triangle are 2m wide, not long)

a) what is the length of the longest beam that can be carried horizontally around this corner (beam cannot bend) hint: trigonometric

b) if the corridor is 3m high, and the beam no longer needs to be carried horizontally, what is the greatest length to be carried around the corner?

2. ## Ok for part one

you have that $V=\frac{1}{2}width\cdot{height}300$...now you need to get rid of height and diffentiate volume but first you want to get rid of that pesky height..but since you know that the length of the sides are 30 you can use the rules of 30-60-90 triangles to ascertain that h=15...now you have that $V=\frac{1}{2}Width15\cdot{300}$ I think you can go from there

3. Originally Posted by Cavaliers06
...

4. a corridor is uniformly 2m wide and makes a right angle. (opposite and adjacent sides of the triangle are 2m wide, not long)

a) what is the length of the longest beam that can be carried horizontally around this corner (beam cannot bend) hint: trigonometric

...

1. Make a sketch.

2. From my drawing you see:

$\frac y2 = \frac{y+2}{2+x} ~\implies~ x = \frac4y$

3. The length of the beam is the hypotenuse of a right triangle. Use Pythagorean theorem:

$l^2 = (y+2)^2 + (x+2)^2 ~\buildrel {x = \frac4y}\over \longrightarrow ~ (l(y))^2 = (y+2)^2 + (\frac4y+2)^2$

4. Differentiate (l(y))² to calculate the extreme value. You are looking for the minimum value!

$((l(y))^2)' = 2(y+2) + 2(\frac4y+2) \cdot \left(-\frac4{y^2}\right)$

5. Calculate

$2(y+2) + 2(\frac4y+2) \cdot \left(-\frac4{y^2}\right) = 0$

You'll get 2 solutions: y = -2 or y = 2

6. If y = 2 then x = 2 too and $l = \sqrt{32}$

The negative length for y isn't very plausibel.

4. For #4a:

There is a formula you can use to find this length lickety-split.

$\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\fra c{3}{2}}$

Where a and b are the widths of the hallways. In this case, a=b=2.

But, to do it by the calc way, we can use similar triangles and noting that x+y=L.

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

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

$L=x+\frac{2x}{\sqrt{x^{2}-4}}$

$\frac{dL}{dx}=1-\frac{8}{(x^{2}-4)^{\frac{3}{2}}}$

Now, set to 0 and solve for x, we find $2\sqrt{2}$

By subbing back into y, we find $y=2\sqrt{2}$

Therefore, the length of the beam is $L=4\sqrt{2}\approx{5.66}$

Let's check it with the formula:

$\left(2^{\frac{2}{3}}+2^{\frac{2}{3}}\right)^{\fra c{3}{2}}\approx{5.66}$

Check.

5. Originally Posted by Cavaliers06

1. A sm long trough is in the shape of an isosceles triangular prism with dimensions 30cm for both the sides of the triangle's face and the length is 300cm. I need to find the width of the triangular prism that will maximize the volume of thr trough. Please show all steps. Recall: V=(A x-section) L

2. Two buildings, A and B, in a school need to be connected with a fibre optic cable. Building A is 70m (up/north) from a roadway and B is 200m down/along the roadway. The cable must be laid underground across the playing field, but along the roadway it can ve constructed above ground. If the cost of it underground is $1000/m while above ground it is$500/m, where should point C be located to minimize the total cost of laying the cable? (ADC is a right angle triangle, and C is a striaght line from B)

3. A future shop store sells on average 20 ipods per week at a price of $200 per ipod. In any given week, the store will have a maximum of 40 ipods for sale. Market research shows that for each$5 increase in price there will be 2 less sales per week, but also for each $5 decrease there will be 2 more sales per week. The cost of the ipods for the store is$100 per ipod. What should the sale price of each ipod be to maximize profit? What is the maximum profit? recall: p(x)= r(x)-c(x)

4. a corridor is uniformly 2m wide and makes a right angle. (opposite and adjacent sides of the triangle are 2m wide, not long)

a) what is the length of the longest beam that can be carried horizontally around this corner (beam cannot bend) hint: trigonometric

b) if the corridor is 3m high, and the beam no longer needs to be carried horizontally, what is the greatest length to be carried around the corner?

Please post one question per thread, that way it is easier for us to see if you have been helped on each question.

It will also make the forum easier to follow.

RonL

6. Originally Posted by Cavaliers06
...

2. Two buildings, A and B, in a school need to be connected with a fibre optic cable. Building A is 70m (up/north) from a roadway and B is 200m down/along the roadway. The cable must be laid underground across the playing field, but along the roadway it can ve constructed above ground. If the cost of it underground is $1000/m while above ground it is$500/m, where should point C be located to minimize the total cost of laying the cable? (ADC is a right angle triangle, and C is a striaght line from B)

...
1. Draw a sketch.

2. The length of the cable is x+y with $y = \sqrt{70^2+(200-x)^2}$

3. The costs for the cable are:

$c = 500x + 1000y$ Plug in the term for y into this equation:

$c(x)=500x + 1000 \sqrt{70^2+(200-x)^2}$

4. Calculate c'(x) = 0

I've got: $x = 200-\frac{70}3 \sqrt{3} \approx 159.585$