Respuesta :
Answer:
Height: 3/2 inches
Length: 12 inches
Width: 4 inches
Step-by-step explanation:
Let x is the side length of the square
The height of the box by cutting squares off :x
- The new length of the cardboard = 15 -2x (because we cut from 4 corners)
- The new width of the cardboard = 7 -2x (because we cut from 4 corners)
The new volume of it is:
V = (15 -2x) (7 -2x) x
<=> V = [tex]4x^3 -44x^2 +105x[/tex]
To maximum volume, we use the first derivative of the volume
[tex]\dfrac{dV}{dx} = 0[/tex]
<=> [tex]12x^2-88x+105=0[/tex]
<=> [tex]\left(2x-3\right)\left(6x-35\right) = 0[/tex]
<=> 2x -3 = 0 or 6x -35 = 0
<=> x = 3/2 or x = 35/6
To determine which value of x gives a maximum, we evaluate
[tex]\dfrac{d^2V}{dx^2}[/tex] = 24x -88
- If x = 3/2, we have:
[tex]\dfrac{d^2V}{dx^2}[/tex]= 24(3/2) -88 = -52
- If x = 35/6, we have:
[tex]\dfrac{d^2V}{dx^2}[/tex] = 24(35/6) -88 = 52
We choose x = 3/2 to have the maximum volume because the value of x that gives a negative value is maximum.
So the dimensions (in inches) of the box is:
Height: 3/2 inches
Length: 15-2(3/2) = 12 inches
Width: 7 - 2(3/2) = 4 inches
To answer this question, it is necessary to use applications of derivatives to find the maximum and or minimum of a function.
Solution is:
Dimensions:
L = 12 in
W = 4 in
h = 1.5 in
V(max) = = 72 in³
Let´s call "x" the side of the identical squares to cut from the edges of the cardboard, then as the cutting is at the four edges we have:
L ( the length of the box is ) = 15 - 2×x in and the wide
W ( the wide of the box) = 7 - 2×x in
h ( the height f the box ) = x in
The volume of the box is: V(b)
V(b) = L × W × h
V(b) = ( 15 - 2×x ) × ( 7 - 2×x ) × x
Volume as a function of x, is:
V(x) = ( 105 - 30×x - 14×x + 4×x²) × x
V(x) = 105×x -44×x² + 4×x³
Tacking derivatives on both sides of the equation, we get
V(x)´ = 105 - 88× x + 12×x²
V(x)´ = 0 ⇒ 105 - 88× x + 12×x² = 0
12×x² - 88×x + 105 = 0
The last expression, is a second-degree equation, solving for x
x₁,₂ = [ 88 ± √7744 - 5040 ] / 24
x₁ = (88 + 52)/24 x₁ = 5.83 we dismiss such value since does not bring a feasible solution
x₂ [ 88 - 52 ]/24 x₂ = 1.5 in
Then the side to cut from de edges is x = 1.5 in, and the sides of the box are:
L = 15 - (1.5)×2 L = 12 in
W = 7 - 2× (1.5) W = 4 in
h = 1.5 in
V(max) = 12×4×1.5
V(max) = 72 in³
To check the maximum, we get the second derivative, if the second derivative V(x)´´ < 0 the function V(x) has a maximum for x = 1.5
V(x)´´ = - 88 + 12 × x ⇒ if x = 1.5 V(x) ´´ = -70 then V(x)´´ < 0
Then V(x) has a maximum for x = 1.5
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