Question

A rectangular storage container with an open top is to have a volume of 10 m^3. The length of this base is twice the width. Material for the base costs $5 per square meter. Material for the sides costs $3 per square meter. Find the cost of materials for the cheapest such container.

Answer 1

what have you tried so far

Answer 2

I don't even know how to approach it

Answer 3

this is a word problem, interpret the problem and always draw a rough diagram even if it is not necessary

Answer 4

next look at the given info :

\(lwh = 10\)
\(l = 2w\)

Answer 5

set up an equation for the cost and minimize it

Answer 6

see if u can write an expression for cost of making the total container

Answer 7

`Material for the base costs $5 per square meter. `

Area of base = \(lw \)
so can we say the cost for making base is \(5lw\) ?

Answer 8

I am so confused right now I'm sorry

Answer 9

its okay, word problems confuse me too
go through the question again and see if the problem makes sense first

Answer 10

I am just confused with the money aspect of it

Answer 11

ikr, thats the most tricky part
lets see

Answer 12

did you get why the cost of making base equals \(5*lw\) ?

Answer 13

Yes I got that part

Answer 14

then we are good :) lets keep going

Answer 15

lets find the cost of making sides

Answer 16

`Material for the sides costs $3 per square meter. `

lateral area of box = (perimeter of base)*height
= \(2(l+w)*h\)

multiply it by 3 to get the total cost : \(3 *2(l+w)*h\)

Answer 17

so far we have this :

cost of making base = \(5lw\)
cost of making sides = \(3*2(l+w)*h\)

Answer 18

So our cost expression for total cost of making the box would be :

\[C = 5lw + 3*2(l+w)*h\]

Answer 19

see if that makes sense so far

Answer 20

It does.

Answer 21

good, we are not done yet
we need to find the minimum value of that Cost function

Answer 22

you must be knowing how to minimize a function having single variable

Answer 23

but in our cost function we have 3 variables,
what to do

Answer 24

somehow we need to eliminate 2 variables

Answer 25

l can equal 2w right?

Answer 26

Yes! we can use the given info

replace \(l\) by \(2w\)

Answer 27

\[C = 5lw + 3*2(l+w)*h\]

becomes

\[C = 5(2w)w + 3*2(2w+w)*h\]

Answer 28

still we have 2 variables
we need to eliminate h somehow..

Answer 29

I have no idea with that one.. Sorry

Answer 30

use the other given info

Answer 31

\(lwh = 10\)

isolate h from here and plug it in ur cost function

Answer 32

okay so \[c=5(2w)w+3*2(2w+w)*10/2w^2\]

Answer 33

Excellent! simplify a bit before starting the calculus part

Answer 34

simiplifying should give you \[C=10w^2+\frac{90}{w}\]

check if it is correct ^

Answer 35

That is what I got too

Answer 36

great! go ahead and minimize it

Answer 37

familiar with the process ?
taking the first derivative and setting it equal to 0..

Answer 38

Yes I got that part. It's the easiest

Answer 39

I agree, calculus is the easiest part :)
algebra sucks though

Answer 40

wolfram gives a weird number for cost
http://www.wolframalpha.com/input/?i=minimize+10w%5E2+%2B+90%2Fw

Answer 41

\[\large \text{minimum cost} = 45 \sqrt[3]{6}\]

Answer 42

I got \[\sqrt[3]{9/20}\]

Answer 43

you got that for \(w\) right ?

Answer 44

Yes

Answer 45

you should get \(\large w = \sqrt[3]{\frac{9}{2}}\)

Answer 46

check ur work again

Answer 47

I found out what I did wrong