Asked by Finneas
The given figure is made up of isosceles triangle on top of a rectangle. If the perimeter is 48 m, determine the exact value (no decimal answer) of x that produces a maximum area.
(The figure looks like a pentagon, where the bottom rectangle has visible dimensions of y by x by y and the isosceles triangle on top has the dimesions of 1.3x by x by 1.3x)
(The figure looks like a pentagon, where the bottom rectangle has visible dimensions of y by x by y and the isosceles triangle on top has the dimesions of 1.3x by x by 1.3x)
Answers
Answered by
R_scott
perimeter ... 2 y + 3.6 x = 48 ... y + 1.8 x = 24 ... y = 24 - 1.8 x
area ... (x * y) + (.5 x * 1.2 x)
substituting ... 24 x - 1.8 x^2 + .6 x^2 ... 24 x - 1.2 x^2
max is on the axis of symmetry ... xmax = -24 / (2 * -1.2)
area ... (x * y) + (.5 x * 1.2 x)
substituting ... 24 x - 1.8 x^2 + .6 x^2 ... 24 x - 1.2 x^2
max is on the axis of symmetry ... xmax = -24 / (2 * -1.2)
Answered by
oobleck
from the dimensions you have given, it appears that
x + 2y + 2.6x = 48
so the area is
A = xy + 1/2 x√((1.3x)^2 - (x/2)^2)
A = x(24-1.8x) + 1/2 x^2√(1.69 - 0.25) = -1.2x^2 + 24x
so to find the maximum area, you need dA/dx = 0
dA/dx = -2.4x + 24
so x = 10
But is this the true maximal area?
What if the triangle is taller or shorter than 1.2x?
x + 2y + 2.6x = 48
so the area is
A = xy + 1/2 x√((1.3x)^2 - (x/2)^2)
A = x(24-1.8x) + 1/2 x^2√(1.69 - 0.25) = -1.2x^2 + 24x
so to find the maximum area, you need dA/dx = 0
dA/dx = -2.4x + 24
so x = 10
But is this the true maximal area?
What if the triangle is taller or shorter than 1.2x?
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