Respuesta :
Answer: The side of the square will be 3 cm.
Step-by-step explanation:
When squares of equal size of side x, are cut from the four corners, we are left with the following dimensions for the cuboid box.
l= 30-2x
b = 14-2x
h= x
Hence, the volume of the cuboid will be ,
V = lbh
= (30-2x)(14-2x)x
Since this volume should be the largest possible, then differentiating volume with x should be equal to zero.
Therefore, [tex]\frac{dV}{dx} = 0[/tex]
[tex]\frac{dV}{dx} =-2(14-2x)x -2(30-2x)x+(30-2x)(14-2x)=0[/tex]
solving we get, x = 3, x = [tex]\frac{35}{3}[/tex]
We can omit x = [tex]\frac{35}{3}[/tex] because 14 - 2x < 0 , which is not possible.
Therefore, x=3.
The cardboard dimensions of 14 by 30 in. require a squares of size 3 in.
side length to be cut from the corners to give the largest volume.
Response:
- The size of the square that gives the largest possible volume has a side length of 3
What method can be used to calculate the correct size of the cut out square?
Length of the cardboard = 30 in.
Width of the cardboard = 14 in.
Let x represent the side length of the cut square, we have;
Length of the cube = 30 - 2·x
Width of the cube = 14 - 2·x
Height of the cube = x
Therefore;
Volume of the cube, V = (30 - 2·x)·(14 - 2·x)·x = 4·x³ - 88·x² + 420·x
At the maximum volume, we have;
- [tex]\dfrac{dV}{dx} = \dfrac{d}{dx} \left(4 \cdot x^3 - 88 \cdot x^2 + 420 \cdot x \right) = \mathbf{12\cdot x^2 - 176 \cdot x + 420} = 0[/tex]
Which gives;
4·(3·x - 35)·(x - 3) = 0
x = 3, or x = [tex]\frac{35}{3}[/tex]
When x = 3, we have;
V = 4·3³ - 88·3² + 420·3 = 573
When x = [tex]\mathbf{\frac{35}{3}}[/tex], we have;
[tex]4 \cdot \left(\dfrac{35}{3} \right) ^3 - 88 \cdot \left(\dfrac{35}{3} \right)^2 + 420 \cdot \left(\dfrac{35}{3} \right) \approx -725.93[/tex]
Therefore;
- The largest possible volume is obtained when the size of the side length of the square is x = 3
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