Respuesta :
Answer:
side of the square x = 2.43 in
Area of the square x² = 5.91 in²
Step-by-step explanation:
The piece of cardboard is
20 in * 12 in (rectangle) we are going to cut four squares each in a corner of that piece then
V(b) = ( 12 - 2x )* ( 20 - 2x) *x
V(b) = ( 240 - 24x - 40x + 4x² ) *x V(b) = ( 240 - 64x + 4x²)*x
V(b) = 240x - 64x² + 4x³
Taking derivatives both sides of equation
V´(b) = 240 - 128x + 12x²
Then V´(b) = 0 240 - 128x + 12x² = 0
A second degree equation solving we have
12x² -128x + 240 = 0 ⇒ 3x² - 32x + 60
x₁,₂ = [ 32 ± √(32)² - 720] / 6 ⇒ x₁,₂ = [ 32 ± 17.44]/6
x₁ = 8.24 in we dismiss this value since 2 times this value is bigger than one side which is not possible
x₂ = 2.43 in then the square is (2.43)² = 5.91 in²
The side length of the square cut out from the 12 inches by 20 inches
cardboard to give the largest volume is approximately 2.43 inches.
How can the correct size of square to cut out be calculated?
The dimensions of the cardboard = 12 inches by 20 inches
Let x represent the size of the square cutout from the corners, we have;
The volume of the box, V = (12 - 2·x)(20-2·x)·x = 4·x³ - 64·x² + 240·x
At the largest volume, we have;
[tex]\dfrac{dV}{dx} = \mathbf{\dfrac{d}{dx} \left(4 \cdot x^3 - 64 \cdot x^2 + 240 \cdot x\right)} = 12\cdot x^2 - 128\cdot x + 240 = 0[/tex]
(12·x² - 128·x + 240) = 0
Which gives;
[tex]x = \dfrac{-(-128) \pm\sqrt{(-128)^2 - 4 \times 12 \times 240} }{2 \times 12} = \mathbf{ \dfrac{128 \pm 16 \cdot \sqrt{19} }{24}}[/tex]
x ≈ 2.43 or x ≈ 8.24
The possible value is x ≈ 2.43 inches
The side length of the square that gives the largest volume of the open box is x ≈ 2.43 inches
Learn more about largest (maximum) value of a function here:
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