The design of a building that has a square pyramid roof as a roof is shown. The cost of material for the outside of the building and for the roof

ranges from $25 per square foot to $50 per square foot. The budget for this material is $500,000. The rectangular front of the building has a

length twice as long as its height. The slant height of the roof is the same as the height of the rectangular front of the building.

What is the maximum possible length of the rectangular front of the building to the nearest foot?

feet

The maximum possible length of the rectangular front of the building is

A. 164

B. 41

C. 82

D. 29

Question

24-07-2022
Mathematics

contestada

The design of a building that has a square pyramid roof as a roof is shown. The cost of material for the outside of the building and for the roof

ranges from $25 per square foot to $50 per square foot. The budget for this material is $500,000. The rectangular front of the building has a

length twice as long as its height. The slant height of the roof is the same as the height of the rectangular front of the building.

What is the maximum possible length of the rectangular front of the building to the nearest foot?

feet

The maximum possible length of the rectangular front of the building is

A. 164

B. 41

C. 82

D. 29

Respuesta :

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MrRoyal MrRoyal · Answer 1 · 2022-07-27T13:36:59+02:00

The maximum possible length of the rectangular front of the building is 23 feet

How to determine the maximum possible length?

The complete question is attached

Let the length of the rectangular front be x and the height be y.

So, we have:

x = 2y

The building has 4 congruent sides.

So, the area of the 4 sides is

A = 4 * (x * y)

This gives

A = 4 * (x * 2x)

Evaluate

A = 8x²

For the triangular roof, we have:

Slant height, l = y

Base, b = x

So, the area of the 4 triangular faces is

A = 0.5 * 4 * xy

This gives

A = 2xy

Recall that:

x = 2y

Make y the subject

y = 1/2x

So, we have:

A = 2x * 1/2x

A = x²

The cost of designing the buildings is

C = 25 * 8x² + 50 * x²

C = 200x² + 50x²

C = 250x²

This gives

250x² = 500000

Divide both sides by 250

x² = 2000

Square both sides

x = 45

Recall that:

y = 1/2x

This gives

y = 1/2 * 45

y = 23

Hence, the maximum possible length of the rectangular front of the building is 23 feet