# Thread: Help getting started with a couple optimizations

1. ## Help getting started with a couple optimizations

Sometimes I can do these... sometimes I look at them and don't know where to begin.

If someone could give me a couple tips to get these few started, I can do the rest:

1. A box with a square base and no top must have a volume of 10 000 cm3. If the smallest dimension in any direction is 5 cm, then determine the dimensions of the box that minimize the amount of material used.

2. A fence is 1.5 m high and is 1 m from a wall. A ladder must start from the ground, touch the top of the fence, and rest somewhere on the wall. Find the minimum length of such a ladder.

3. The motion of a particle is given by s(t) = 5cos(2t + pi/4). What are the maximum values of the displacement, the velocity, and the acceleration?

2. Originally Posted by NAPA55
Sometimes I can do these... sometimes I look at them and don't know where to begin.

If someone could give me a couple tips to get these few started, I can do the rest:

1. A box with a square base and no top must have a volume of 10 000 cm3. If the smallest dimension in any direction is 5 cm, then determine the dimensions of the box that minimize the amount of material used.

2. A fence is 1.5 m high and is 1 m from a wall. A ladder must start from the ground, touch the top of the fence, and rest somewhere on the wall. Find the minimum length of such a ladder.

3. The motion of a particle is given by s(t) = 5cos(2t + pi/4). What are the maximum values of the displacement, the velocity, and the acceleration?
For the first one the surface area is given by the formula

$\displaystyle \underbrace{x^2}_{bottom}+\underbrace{4xy}_{sides} =10,000 \mbox{ and } V=x^2y$

Solve the first for y and sub into the second and minimize the function

$\displaystyle V=2500x-\frac{x^3}{4}$

3. ## Heres three

the first two are fairly simple $\displaystyle f(t)=5cos\bigg(2t+\frac{\pi}{4}\bigg)$...then $\displaystyle v(t)=f'(t)=-5sin\bigg(2t+\frac{\pi}{4}\bigg)\cdot{2}$...set that equal to zero you get...wait there is no boundary or artificial cut off...well I am just going to assume it is $\displaystyle (0,2\pi)$ the values are $\displaystyle x=\frac{3\pi}{8},x=\frac{7\pi}{8}.x=\frac{11\pi}{8 },x=5.89$ next put those values in the second derivative...the negative values are maxs and the minimums are negative...repeat for the other two with the derviative...solving and putting that in its derivative

4. ## similar triangles

We need to use similar triangles to find d in the diagram below.(sorry without the buttons I don't know how to add it in the middle of a post)

$\displaystyle \frac{d}{1.5}=\frac{x+1}{x} \iff d=\frac{3(x+1)}{2x}$

Since we want the smallest length of the latter we need to know the length of the hypotenuse. By the pythagorean theorem we get

$\displaystyle h=\sqrt{\left( \frac{3(x+1)}{2x}\right)^2+(x+1)^2}$

You just need to minimize from here.

Good luck.

5. Thank you both!