An open box is to be made from cutting squares of side 's' from each corner of a piece of cardboard that is 35" by 40".
(a) Write an expression for the volume, 'V', of the box in terms of 's'.
(b) Draw a graph of V(s).
(c) State the domain and range of V(s).
(d) Find the value of 's' that will give the maximum volume. What is the maximum volume?
(e) What realistic value(s) of 's' will generate a volume of 1225 cubic units?
length = 40-2s
width = 35-2s
height = s
a) V = s(40-2s)(35-2s)
= ..
= 1400s - 150s^2 + 4s^3
b) standard shape of a cubic, with x-intercepts of
0 , 20, and 17.5
c) make special notice of where V is above and below the x-axis
d) dV/ds = 1400 - 300s + 12s^2
= 0 for a max of V
divide by 4 ...
3s^2 - 75s + 350 = 0
s = (75 ± √1425)/6
= appr 18.8 or appr 6.2
clearly the 18.8 would produce a negative width, so we reject that
accept s = 6.2 to yield a max volume of appr 3867 cubic inches
to get a volume of 1225 we set
4s^3 - 150s^2 + 1400s = 1225
4s^3 - 150s^2 + 1400s - 1225 = 0 gives a solution of
s = .974 , 13.9 and 22.6
by Wolfram:
http://www.wolframalpha.com/input/?i=4s%5E3+-+150s%5E2+%2B+1400s+-+1225%3D0
both .974 and 13.9 yield our needed result, while 22.6 would produce a negative length and width, thus we must reject it.
(a) The volume of the box can be expressed as V = s(35-2s)(40-2s).
(b) The graph of V(s) will need to be plotted based on the expression V = s(35-2s)(40-2s). It is a cubic function and can be graphed to show how the volume changes with different values of s.
(c) The domain of V(s) is the set of all possible values of s for which the expression is defined. In this case, s should be between 0 and half the length of the shorter side of the cardboard, so 0 ≤ s ≤ 17.5. The range of V(s) is the set of all possible volumes that can be attained by varying the value of s. Since volume cannot be negative, the range is V ≥ 0.
(d) To find the value of s that will give the maximum volume, we can take the derivative of the expression V = s(35-2s)(40-2s) with respect to s and set it equal to zero:
dV/ds = 35(40-2s) + s(-4)(40-2s) + s(35-2s)(-2) = 0
Simplifying the equation, we get:
80s - 4s^2 - 70s + 8s^2 - 70s + 4s^2 = 0
Combining like terms, we have:
-2s^2 +40s = 0
Factoring out -2s, we get:
-2s(s - 20) = 0
Setting each factor equal to zero, we find:
-2s = 0 or s - 20 = 0
Solving for s, we have:
s = 0 or s = 20
Since s cannot be zero in this case, we have s = 20 as the value that gives the maximum volume.
To find the maximum volume, substitute s = 20 into the original volume expression:
V = 20(35-2(20))(40-2(20))
Simplifying, we get:
V = 20(35-40)(40-40)
V = 20(15)(0)
V = 0 cubic units
Therefore, the maximum volume is 0 cubic units.
(e) To find the realistic values of s that will generate a volume of 1225 cubic units, we can set the volume expression equal to 1225 and solve for s:
1225 = s(35-2s)(40-2s)
This is a cubic equation that can be solved either algebraically or graphically. Solving this cubic equation will give the realistic values of s that generate a volume of 1225 cubic units.
(a) To find the volume of the open box, we need to determine the dimensions of the box, which are determined by the side length of the squares cut from each corner. Let's denote the side length of the squares as 's'.
Since squares are cut from each corner, the length of the box will be reduced by 2s, and the width will be reduced by 2s as well. Therefore, the length and width of the box will be (35 - 2s) and (40 - 2s) respectively. The height of the box will be equal to 's'.
So, the expression for the volume of the box is:
V = (length)(width)(height)
V = (35 - 2s)(40 - 2s)(s)
(b) To draw a graph of V(s), we can plot the volume (V) on the y-axis and the side length of the squares (s) on the x-axis. We can assign various values to 's' and calculate the corresponding volumes using the expression from part (a). Plotting the points will give us the graph.
(c) The domain of V(s) represents the valid values for 's,' which in this case would be the side length of the squares that can be cut from the cardboard without causing any issue. The range of V(s) represents the possible volumes of the box that can be obtained for different values of 's.'
In this problem, the domain for 's' would be 0 ≤ s ≤ 17.5 since we cannot cut squares larger than half the length or width of the cardboard. The range for V(s) would be V ≥ 0, as volume cannot be negative.
(d) To find the value of 's' that gives the maximum volume, we need to find the maximum point on the graph of V(s).
To do this, we can use calculus by taking the derivative of V(s) with respect to 's' and setting it equal to zero to find the critical points. Then, we evaluate the second derivative to determine if the critical points are maximum or minimum points.
V(s) = (35 - 2s)(40 - 2s)(s)
Differentiating V(s) with respect to 's':
dV/ds = -(80 - 4s)(35 - 2s) + (35 - 2s)(40 - 2s)
Setting the derivative equal to zero and solving for 's':
-(80 - 4s)(35 - 2s) + (35 - 2s)(40 - 2s) = 0
By solving this equation, you can find the value(s) of 's' that give the maximum volume. Plug these values back into the expression for V(s) to calculate the maximum volume.
(e) To find the realistic value(s) of 's' that will generate a volume of 1225 cubic units, we can set the volume expression V(s) equal to 1225 and solve for 's':
(35 - 2s)(40 - 2s)(s) = 1225
By solving this equation, we can find the value(s) of 's' that will give a volume of 1225 cubic units.