Question

Max-Min problem please help

Answer 1

a rectangular warehouse is to have 3300 square feet of floor area and is to be divided into two rectangular rooms by interior wall. Cost per running foot is $125 for exterior walls and $80 for the interior wall.
(a) What dimensions will minimize total wall cost?
(b) What is the minimum cost?

Answer 2

did you draw a picture and get an equation?

Answer 3

S=xy , f(x,y)=125(2x+2y)+80y=250x330y
f(x,3300/y)=250x+330(3300/x)
(250x+330(3300/x))'=250-1089000/x^2
(250x+330(3300/x))''=2178000/x^3

Answer 4

i dont get it after this

Answer 5

cost = 125(2x+2y)+80x
y = 3300/x

cost \(= 125(2x+2(\frac{3300}{x}))+80x \)

Answer 6

to minimize the cost, take the derivative, set it equal to 0 and solve for x.

that will give you the dimension of the inner wall.
use it in the area equation to find the other dimension, and use it in the cost equation to find the minimum cost.

Answer 7

i did that first derivative 250-1089000/x^2

Answer 8

x=66

Answer 9

no, it should be 250(1-6600/x^2)+80 =0

multiply thru by x^2 to get
250(x^2-6600) + 80x^2 = 0

=> 330x^2 - 250(6600) = 0 => 330(x^2-5000) = 0

=> x^2 - 5000 = 0 so \(x=\pm\sqrt{5000}\)

Answer 10

only the positive root will work.

=> \(y=3300/\sqrt{5000}= 46.669...\)

Answer 11

ok this is for part A right

Answer 12

and for part B i just have to plug in x and y into the cost equation

Answer 13

@pgpilot326

Answer 14

you got it

Answer 15

actually, the cost equation is already in terms of x so it's all you need.

Answer 16

you good?

Answer 17

is it 35001.78

Answer 18

i don't know, i didn't compute