A box with a square base with no top has a surface area of 108 square feet Find the dimensions that will maximize the volume Remember you want to maximize volume NOT surface area?

Answer 1

So first lets figure out what we know (and what it tells us): We are dealing with a box (assuming this is a normal box, the height of all the sides will be the same) Box has a square base (the width of all the sides will be the same) Box has no top (we only need to account for 5 sides when calculating surface area) Surface area = 108 ft2 Now let us write us equations from this information: Area of Base = W2 (we know the base is square, so L = W for the base; also because it is a box with a square base the width of each side of the box will be equal as well) Area of Side = L * W (in this case length can be thought of as the height of the box side) Total Area of 4 Sides and Base (Surface Area) = 4(L*W) + W2 = 108 Remember we are trying to maximize volume of the box. So let's turn to that equation: Volume = Length * Width * Height The length and width of the box is equal to the area of the base, which we know from above is equal to W2. We need to manipulate the above surface area equation to get the value of height in terms of W. 4(L*W) + W2 = 108 4(L*W) = -W2 + 108 L = -(1/4)W +27/W Now substitute into the volume equation: Volume = W2 * [-(1/4)W +27/W] Volume = (-1/4)W3 +27W Now that we have the volume in terms of one variable we can take the derivitive. Remember that the derivitive will tell us the rate at which volume is changing given different values for W. dV/dW = -(3/4)W2 + 27 By setting the derivitive equal to zero we will get an a minimum or a maximum (a critical point). -(3/4)W2 + 27 = 0 -(3/4)W2 = -27 W2 = 36 W = 6, -6 The width cannot be a negative number so we can discard that answer. In the case that both were positive numbers (and in this case just as a check), we should test to see whether our critical point is a minimum or a maximum. To do this take the second derivitive of the original equation: V = (-1/4)W3 +27W V' = -(3/4)W2 + 27 V'' = -(6/4)W At W = 6 the second derivitive is negative. Using the rule for the second derivitive test, if f''(x) < 0 then there is a local maximum at x. This confirms that our volume equation is maximized at W = 6 Unfortunately, we're not quite done. We found the width now we need the other dimensions. For that we can use our surface area equation to find L (the height) 4(L*W) + W2 = 108 4(L*6) + 62 = 108 24L + 36 = 108 24L = 72 L = 3 Now we're done. The dimensions of the box with the maximum volume given the information provided is 6ft x 6ft x 3ft.