Question

Optimization help? what are the steps to solving this
A closed rectangular container with square base is to have a volume of 2250 in^3. The material for the top and bottom of the container will cost $2 per in^2,and the material for the sides will cost $3 per in^2.Find the dimensions of the container of least cost.

Answer 1

what do you have so far

Answer 2

I just started so v=2xy=2250in^3

Answer 3

it would be x^2y sorry

Answer 4

so far, so good

yeah V = x^2*y = 2250

Answer 5

what is the surface area of the top?

Answer 6

x^2

Answer 7

you have 2 bases like that: top and bottom

so we have 2x^2 for the combined areas

Answer 8

we are told "The material for the top and bottom of the container will cost $2 per in^2"

so how much will the top and bottom cost (in terms of x)?

Answer 9

My mind is blanking, I have no idea

Answer 10

how about just the top only

Answer 11

if the x is throwing you off, imagine x is some fixed number (like 3 or something)

then try to calculate the area of the top and then calculate how much it will cost to cover the top

Answer 12

So it would be 2x^2 * 2

Answer 13

for the cost

Answer 14

if it was 3, the cost would be $18

Answer 15

good, the total cost of top and bottom is 4x^2

we'll keep things in terms of x & y for now

Answer 16

now we move onto the sides

Answer 17

what is the area of one side (in terms of x and y)?

Answer 18

the cost of the sides is 4xy * 3?

Answer 19

yes or 12xy

Answer 20

so the total cost to cover this prism is 4x^2 + 12xy

Answer 21

So then how do I find the minimum price?

Answer 22

recall that above you found x^2*y = 2250

solve for y to get

x^2*y = 2250

y = 2250/(x^2)

then you can plug this into 4x^2 + 12xy to get

4x^2 + 12xy

4x^2 + 12x(2250/(x^2))

4x^2 + 27000/x

Answer 23

so the cost function is now f(x) = 4x^2 + 27000/x based on the condition that x^2*y = 2250

Answer 24

minimizing f(x) will give you the min cost

Answer 25

To find minimum, do I set the derivative of f(x) equal to zero, find the critical values, then plug them into f(x)?

Answer 26

before you plug them into f(x), you need to do the first derivative test

to see if you actually have a min, max or saddle point

Answer 27

but yeah

step 1) find f ' (x)

step 2) solve f ' (x) = 0 for x to get the critical values

step 3) use the first derivative test to determine the nature of each critical value

step 4) once you have the x value where the local min is located, plug it into f(x) to get the min cost

Answer 28

thank you very much

Answer 29

you're welcome

Answer 30

@jim_thompson5910 , how do I do the first derivative test if I was not given an interval to test on?

Answer 31

well x is the side length, so x >= 0 because you can't have negative side lengths

Answer 32

what critical values did you get

Answer 33

15

Answer 34

I dont know if I did it correctly

Answer 35

set up a number line with the critical value on it

Answer 36

then test values to the left and right of 15

say 14 and 16



you test them in f ' (x)

Answer 37

you test them in f ' (x) because you want to determine the tangent slopes of f(x) to figure out if f is increasing or decreasing

Answer 38

let me know what you get for f ' (14) and f ' (16)

Answer 39

f(14)=2584
f(16)=2711.5

Answer 40

f PRIME not just f

Answer 41

you test them in f ' (x) because you want to determine the nature of the slopes

Answer 42

yes f'(x) sorry

Answer 43

I got f'(14)= 2584
and f'(16)=2711.5

Answer 44

I'm not getting those values

Answer 45

what did you get for f ' (x) ?

Answer 46

how about f'(14)= 249.76
and f'(16)= 233.47

Answer 47

what did you get for the derivative function

Answer 48

f'(x) = (8x^3 +27,000)/x^2 ??

Answer 49

it should be -27000

Answer 50

f ' (x) = (8x^3 - 27000)/(x^2)

Answer 51

the negative is because 27000/x = 27000x^(-1)

deriving that gives -27000x^(-2) = -27000/(x^2)

Answer 52

So f'(14)= -25.76
and f"(16)= 22.53

Answer 53

much better