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
Read more about areas at:
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