An open rectangular box is to be made from a piece of cardboard 8 in. wide and 15in. long by cutting squares from the corners and folding up the sides. Find the dimensions of the box of the largest volume? Show work pleeease:) thanks
-
Call the length of one of the squares to be cut (the removed part) 's'. When you fold up the box, the height will be s, and the width and length will be:
W = 8-2s
L = 15-2s
Draw a picture if those equations are unclear. The equation for the volume then is
V = W*L*H = (8-2s)*(15-2s)*s
V = (120 - 46s + 4s^2)*s
V = 120s - 46s^2 + 4s^3
taking the derivative of the volume with respect to our unknown length and setting equal to zero, we have
dV/ds = 120 - 92s + 12s^2 = 0
Rearranging for clarity, we have
12s^2 - 92s + 120 = 0
Using the quadratic equation, two possible solutions are
s1 = [-b+sqrt(b^2-4ac)]/2a
s2 = [-b-sqrt(b^2-4ac)]/2a
Where a, b, and c are the coefficients of the equation (12, -92, and 120 respectively). This yields
s1 = 6
s2 = 5/3
To determine which one is actually a maximum, we take the second derivative of volume to see whether the rate of change is increasing or decreasing at these points (if the second derivative is negative, then it is a local maximum because the rate of change is decreasing, i.e., the function is 'concave down'). Calling the second derivative V`` for clarity, we have
V`` = 24s-92
Plugging in our possible maxima, we have
24*(s1)-92
24*(6)-92 = 52
for the other point:
24*(s2)-92
24*(5/3)-92 = -52
Therefore, s2 is the local maximum. (We know that s1 cannot be the answer because that would give you a negative width! Also, based on the rule above, s1 is a local minimum because the second derivative is positive at that point.) The corresponding volume is:
V = (8-2*5/3)(15-2*5/3)*5/3
V = 90.741 cu. in.
W = 8-2s
L = 15-2s
Draw a picture if those equations are unclear. The equation for the volume then is
V = W*L*H = (8-2s)*(15-2s)*s
V = (120 - 46s + 4s^2)*s
V = 120s - 46s^2 + 4s^3
taking the derivative of the volume with respect to our unknown length and setting equal to zero, we have
dV/ds = 120 - 92s + 12s^2 = 0
Rearranging for clarity, we have
12s^2 - 92s + 120 = 0
Using the quadratic equation, two possible solutions are
s1 = [-b+sqrt(b^2-4ac)]/2a
s2 = [-b-sqrt(b^2-4ac)]/2a
Where a, b, and c are the coefficients of the equation (12, -92, and 120 respectively). This yields
s1 = 6
s2 = 5/3
To determine which one is actually a maximum, we take the second derivative of volume to see whether the rate of change is increasing or decreasing at these points (if the second derivative is negative, then it is a local maximum because the rate of change is decreasing, i.e., the function is 'concave down'). Calling the second derivative V`` for clarity, we have
V`` = 24s-92
Plugging in our possible maxima, we have
24*(s1)-92
24*(6)-92 = 52
for the other point:
24*(s2)-92
24*(5/3)-92 = -52
Therefore, s2 is the local maximum. (We know that s1 cannot be the answer because that would give you a negative width! Also, based on the rule above, s1 is a local minimum because the second derivative is positive at that point.) The corresponding volume is:
V = (8-2*5/3)(15-2*5/3)*5/3
V = 90.741 cu. in.
-
Where is your attempt? You here to learn.