A rectangular building is to have a volume of 8000 fr^3. Annual heating and cooling costs will amount to $2/ft^2 for its top, front, and back, and $4/ft^2 for the two end walls. What dimensions of the building would minimize these annual heating and cooling costs?

the dimensions that will minimize the cost are; 25.2 ft × 25.2 ft × 12.6 ft

Step-by-step explanation:

If the length, width and height of the building are represented by a, b and c, then we have:

Volume = abc

We are given volume as 8000 ft³. Thus;

abc = 8000

Top area = ab

Front area = ac

Back area = ac

End walls area = bc

We are told that cooling costs will amount to $2/ft² for its top, front, and back, and $4/ft² for the two end walls.

Thus, cooling cost equation is;

X = 2(ab) + 2(ac) + 2(ac) + 4bc + 4bc

X = 2ab + 4ac + 4bc

Now, from the volume equation where abc = 8000.

c = 8000/ab

Thus;

X = 2ab + 4a(8000/ab) + 4b(8000/ab)

X = 2ab + 32000/b + 32000/a

To minimize, we will find the a and b derivative and set the equation to zero in both cases.

Thus;

dX/da = 2b - 32000/a² = 0

2b = 32000/a² - - - eq(1)

Also;

dX/db = 2a - 32000/b² = 0

2a = 32000/b² - - - (eq 2)

Making a the subject in eq 2 gives;

a = 32000/2b²

a = 16000/b²

Putting 16000/b² for a in eq (1) gives;

2b = 32000/(16000/b²)²

2b = 32000b⁴/256000000

2b = b⁴/8000

Multiply both sides by 8000 to get;

2b × 8000 = b⁴

16000b = b⁴

Divide both sides by b to get;

16000 = b³

b ≈ 25.2 ft

Putting 25.2 for b in a = 16000/b², we have;

a = 16000/25.2²

a ≈ 25.2

Putting a and b = 25.2 in the volume equation, we have;

25.2 × 25.2 × c = 8000

c = 8000/(25.2 × 25.2)

c = 12.6 ft

Thus, the dimensions that will minimize the cost are; 25.2 ft × 25.2 ft × 12.6 ft

b. x- 48 =120

step-by-step explanation:

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