1)Bob wants to create two pens, as shown in the figure. One pen is for a garden

Question

1)Bob wants to create two pens, as shown in the figure. One pen is for a garden and it needs a heavy duty fence to keep out the critters. This heavy duty fence costs $10 per foot. The dog pen shares a side with the garden and has a lighter weight fence on the other three sides that costs $5 per foot. If each pen is to have an area of 480, find the values of x and y that would minimize the total cost of the fencing.

So basicaly the side connecting the two boxes is a heavy fence - I got up to making my two equations but after that it gets super messy and Im getting th ewrong answer multiple times.
1) xy=480

2) 30x+25y is the cost then what?


also i need help on this question

A box with a square base and open top must have a volume of 32,000 cm3. Find the dimensions of the box that minimize the amount of material used.

So i got SA= 4xy+x^2 and 32000=x^2 y
when i pu tthem together and get the derivative i get some crazy number
and this equation -128000x^-2 +2x

Solve with explanations?

Explanation / Answer

Sol :-

C = 10?(2x+2y)+5?(2x+y) = 30x+25y and xy = 480
Answer: x = 16 feet and y = 19.2 feet

B) A box with a square base and open top must have a volume of 32,000 cm3. Find the dimensions of the box that minimize the amount of material used.

Sol:-

First, the restriction is (w^2)h = 32000

Surface area SA = s^2 + 4sh

note that h = 32000 / (s^2)

Replace h with the above expression in the surface area

SA = s^2 + 4s [ 32000 / (s^2) ]

SA = s^2 + 128000 / s

Differentiate

dSA / ds = 2s - 128000 / s^2

The minimun value of SA is when dSA / ds is zero

0 = 2s - 128000 / s^2

2s = 128000 / s^2

2s^3 = 128000

s^3 = 64000

s = 40

h = 32000 / (40)^2

h = 20


s = 40
h = 20

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