Select the correct answer from the drop-down menu.
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

Question

22-03-2022
Mathematics

contestada

Select the correct answer from the drop-down menu.
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

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MrRoyal MrRoyal · Answer 1 · 2022-03-30T17:02:09+02:00

The maximum length of the rectangular front is the highest length of the rectangular front, and the value is 128.1 feet

How to determine the maximum possible length?

Let the dimension of the rectangular front be x and y.

Such that:

y = 2x

So, the area of the rectangular front is:

A = xy

[tex]A = 2x^2[/tex]

The surface area of the pyramid roof is:

A = 4bh

Where:

b = base =25

h = height =18

So, we have:

[tex]A = 4* 25 * 18[/tex]

[tex]A = 1800[/tex]

The total surface area of the figure is:

[tex]T = 1800 +2x^2[/tex]

The range of the cost of material is: $25 per square foot to $50 per square foot.

So, the maximum cost is:

[tex]50 * (1800 +2x^2) = 500000[/tex]

Divide both sides by 50

[tex]1800 +2x^2 = 10000[/tex]

Subtract 1800 from both sides

[tex]2x^2 = 8200[/tex]

Divide both sides by 2

[tex]x^2 = 4100[/tex]

Take the square root of both sides

x = 64.03

Recall that:

y = 2x

So, we have:

y = 2 * 64.03

y = 128.06

Approximate

y = 128.1

Hence, the maximum possible length of the rectangular front of the building is 128.1 feet