Question

ok so Optimization, i just learned this so bear with me.

Find the length and width of a rectangle that has an area of 968 square feet and whose perimeter is a minimum.

I was getting the ones where i maximized correct, but how does my approach differ when minimizing?

Answer 1

do your own homework sir!

Answer 2

eh! i just don't understand why i cant get the minimum problems right :D if someone could please explain, i already know the answer, but i don't understand

Answer 3

Im not always correct with optimization. So, ill let someone else help you. You should seek help from @jim_thompson5910 or @phi ..they know they're stuff and are really helpful with explanations.

Answer 4

A = LW

968 = LW

968/W = L

L = 968/W

-----------------------

P = 2L + 2W

P = 2(968/W) + 2W

P = 1936/W + 2W

P = 1936/W + (2W^2)/W

P = (1936 + 2W^2)/W

You want to minimize this, so derive it with respect to W. Then set the derivative equal to 0 and solve for W. This will give you the critical values, then you can use either the first or second derivative test to see which critical values are the min/max.

Answer 5

hmm i still got the same answer.

Answer 6

what answer are you getting

Answer 7

well my x value is 22squareroot2

Answer 8

that's the value of W that yields the smallest perimeter

Answer 9

use that to find the length

Answer 10

but apparently the answer is 484ft by 2ft

Answer 11

those are one set of dimensions for this rectangle, but they don't produce the smallest perimeter

Answer 12

the perimeter of that 484ft by 2ft rectangle is

P = 2L + 2W

P = 2*484 + 2*2

P = 968 + 4

P = 972

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but the rectangle 242 ft by 4 ft gives you a smaller perimeter (still not the smallest though)

P = 2L + 2W

P = 2*242 + 2*4

P = 484 + 8

P = 492


so the answer can't be 484ft by 2ft just by this counter-example alone

Answer 13

so there's clearly a typo somewhere

Answer 14

:/ ok i'll ask my teacher tomorrow. thanks for all your help!

Answer 15

yw