1. ## Optimization

I am having trouble setting up the following problem.

A rectangular storage container with an open top is to have a volume of 10 m3. The length of this base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs$6 per square meter. Find the cost of materials for the cheapest such container.

I really don't see what equations I could get out of that information.

While I am at it, I have never worked with optimization problems that involve radians.

A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of the sheet on each side through an angle θ. How should θ be chosen so that the gutter will carry the maximum amount of water?

Again I just need someone to point me in the right direction. Namely, how and what the equations should be.

2. Originally Posted by mj11774
I am having trouble setting up the following problem.

A rectangular storage container with an open top is to have a volume of 10 m3. The length of this base is twice the width.
Let x be the width of the base, y the height. Then the length of the base is 2x and the volume= $2x^2y= 10$ so $y= \frac{5}{x^2}$.

Material for the base costs $10 per square meter. The area of the base is $x(2x)= 2x^2$ so the cost of the base is $20x^2$ dollars. Material for the sides costs$6 per square meter. Find the cost of materials for the cheapest such container.
Two of the sides have area $xy= \frac{5x}{x^2}= \frac{5}{x}$. The cost of those two sides is $(2)(6)\frac{5}{x}= \frac{60}{x}$.
Two of the sides have area $(2x)y= \frac{10}{x^2}= \frac{10}{x}$. The cost of those two sides is $2(6)\frac{10}{x}= \frac{120}{x}$.

The total cost of the container is $20x^2+ \frac{60}{x}+ \frac{120}{x}= 20x^2+ \frac{180}{x}$.

That is the function you want to maximize.

I really don't see what equations I could get out of that information.

While I am at it, I have never worked with optimization problems that involve radians.

A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of the sheet on each side through an angle θ. How should θ be chosen so that the gutter will carry the maximum amount of water?

Again I just need someone to point me in the right direction. Namely, how and what the equations should be.
The gutter will "carry the maximum amount of water" when the area of the triangle formed is maximum. If you have an isosceles triangle with sides of length 30/2= 15 cm and vertex angle $\theta$, then you have two right triangles with hypotenuse of length 15 and angle $\theta/2$. The "opposite side", which is at the top of the gutter, is $15 sin(\theta/2)$ so the width of the entire gutter, the "base" of the triangle, is $30 sin(\theta/2)$. The "near side", the altitude of the entire triangle, is $15 cos(\theta/2)$.

Since "Area= (1/2)base*height" for a triangle, the area of the triangle formed by this gutter, and the function you want to maximize, is $225 sin(\theta/2)cos(\theta/2)$.

3. Originally Posted by mj11774
A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of the sheet on each side through an angle θ. How should θ be chosen so that the gutter will carry the maximum amount of water?

Again I just need someone to point me in the right direction. Namely, how and what the equations should be.
The gutter will carry the maximum amount of water when the area of the trapezoid formed is maximum.

$A=\frac{1}{2}h(b_1+b_2)$

$h=10sin{\theta}$

$b_1=10$

$b_2=10+2(10cos{\theta})$

4. ## Thank You

Could not for the life of me figure out the equations. Thank you both for the great help.