Answer question 3&4 please!
1 answer:
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Answer:
V (max) = 2 ft³
and x side of the base is x = 0,5 feet
Step-by-step explanation: See annex ( two different cubes)
We have a square piece of cardboard of 3 inches wide
Let x be lenght of side to cut in each corner
Then the base of the (future cube) is 3 - 2x, and the area is
( 3 - 2x )²
And the height is x Then volume of the cube as a function of x is:
V(x) = ( 3 - 2x )² *x or V(x) = ( 9 + 4x² - 12x )*x
V(x) = 4x³ - 12x² + 9x
Taking derivatives on both sides of the equation
V´(x) = 12x² - 24x + 9
V´(x) = 0 12x² - 24x + 9 = 0 simplifying 4x² - 8x + 3 = 0
Second degree equation solving for x
x₁,₂ = [ 24 ± √( 576) - 432 /24
x₁,₂ = [24 ±√144 ]/24
x₁,₂ = ( 24 ± 12) /24 x₁ = 1.5 feet x₂ = 0,5 feet
Of these two values we have to dismiss x₁ because if x = 1.5 we don´t have a cube ( 0 height )
Then we take x = 0,5 feet
And
V (max) = (2)²*0,5 = 4*0,5
V (max) = 2 ft³
