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Answer:
The width of the box is 6.7 cm
The maximum volume is 148.1 cm³
Step-by-step explanation:
The given parameters of the box Jorge is asked to build are;
The maximum girth of the box = 20 cm
The nature of the sides of the box = 2 square sides and 4 rectangular sides
The side length of square side of the box = w
The length of the rectangular side of the box = l
Therefore, we have;
The girth = 2·w + 2·l = 20 cm
∴ w + l = 20/2 = 10
w + l = 10
l = 10 - w
The volume of the box, V = Area of square side × Length of rectangular side
∴ V = w × w × l = w × w × (10 - w)
V = 10·w² - w³
At the maximum volume, we have;
dV/dw = d(10·w² - w³)/dw = 0
∴ d(10·w² - w³)/dw = 2×10·w - 3·w² = 0
2×10·w - 3·w² = 20·w - 3·w² = 0
20·w - 3·w² = 0 at the maximum volume
w·(20 - 3·w) = 0
∴ w = 0 or w = 20/3 = 6.
Given that 6. > 0, we have;
At the maximum volume, the width of the block, w = 6. cm ≈ 6.7 cm
The maximum volume, , is therefore given when w = 6.
cm = 20/3 cm as follows;
V = 10·w² - w³
= 10·(20/3)² - (20/3)³ = 4000/27 = 148.
The maximum volume, = 148.
cm³ ≈ 148.1 cm³
Using a graphing calculator, also, we have by finding the extremum of the function V = 10·w² - w³, the coordinate of the maximum point is (20/3, 4000/27)
The width of the box is;
6.7 cm
The maximum volume is;
148.1 cm³
Answer:
45 square meters
Step-by-step explanation:
we know that
The area of trapezoid is equal to
where
b1 and b2 are the parallel bases
h is the height of trapezoid (perpendicular distance between the parallel bases)
we have
substitute