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Fantom [35]
2 years ago
12

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 41 in. by 22 in. by

cutting congruent squares from the corners and folding up the sides. Then find the volume.
Mathematics
1 answer:
BlackZzzverrR [31]2 years ago
7 0

Answer:

V(max) = 1872.42 in³

Step-by-step explanation:

Let call  " x " the side of the cut square in each corner, then the sides of rectangular base ( L  and  W )will be

L  =  41 - 2*x       and

W =  22 - 2*x             x = the height of the open box

Volume of the box is

V(b)  =  L*W*x

And volume of the box as a function of x is:

V(x)  =  ( 41 - 2*x ) * ( 22 - 2*x ) * x

V(x) = ( 902 -82*x - 44*x  + 4*x² ) * x   ⇒  V(x) = ( 902 -126*x + 4x²) * x

V(x) =  902*x  - 126*x² +  4x³

Taking derivatives on both sides of the equation we get:

V´(x) = 902  -  252*x + 12 *x²

V´(x) = 12*x²  -  252*x  + 902

V´(x) =  0         ⇒      12*x²  -  252*x  + 902  = 0

A second degree equation solving for x (dividing by 2 )

6*x²  - 126* x  + 451  = 0

x₁,₂  = ( 126 ± √  15876 - 10824 ) / 12

x₁,₂  = (  126  ±  71,08 ) / 12

x₁  =  16 in   ( we dismiss this value since 2x  > than 22 there is not feasible solution for the problem)

x₂ =  54.92/12       ⇒   x₂ =  4,58 in

Then maximum volume is:

V(max) = 31.84 * 12.84 * 4,58

V(max) = 1872.42 in³

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Step-by-step explanation:

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First of all, have to find gradient using the formula above :

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m = (-2-4) / (14-2)

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3 years ago
Find the point on the parabola y^2 = 4x that is closest to the point (2, 8).
guapka [62]

Answer:

(4, 4)

Step-by-step explanation:

There are a couple of ways to go at this:

  1. Write an expression for the distance from a point on the parabola to the given point, then differentiate that and set the derivative to zero.
  2. Find the equation of a normal line to the parabola that goes through the given point.

1. The distance formula tells us for some point (x, y) on the parabola, the distance d satisfies ...

... d² = (x -2)² +(y -8)² . . . . . . . the y in this equation is a function of x

Differentiating with respect to x and setting dd/dx=0, we have ...

... 2d(dd/dx) = 0 = 2(x -2) +2(y -8)(dy/dx)

We can factor 2 from this to get

... 0 = x -2 +(y -8)(dy/dx)

Differentiating the parabola's equation, we find ...

... 2y(dy/dx) = 4

... dy/dx = 2/y

Substituting for x (=y²/4) and dy/dx into our derivative equation above, we get

... 0 = y²/4 -2 +(y -8)(2/y) = y²/4 -16/y

... 64 = y³ . . . . . . multiply by 4y, add 64

... 4 = y . . . . . . . . cube root

... y²/4 = 16/4 = x = 4

_____

2. The derivative above tells us the slope at point (x, y) on the parabola is ...

... dy/dx = 2/y

Then the slope of the normal line at that point is ...

... -1/(dy/dx) = -y/2

The normal line through the point (2, 8) will have equation (in point-slope form) ...

... y - 8 = (-y/2)(x -2)

Substituting for x using the equation of the parabola, we get

... y - 8 = (-y/2)(y²/4 -2)

Multiplying by 8 gives ...

... 8y -64 = -y³ +8y

... y³ = 64 . . . . subtract 8y, multiply by -1

... y = 4 . . . . . . cube root

... x = y²/4 = 4

The point on the parabola that is closest to the point (2, 8) is (4, 4).

4 0
2 years ago
The question is..........
garri49 [273]
10-5=5 and 16-10=6 then you multiply 6x5
the answer is 30 Because you have to multiply L x W 

5 0
3 years ago
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