Suppose an airline policy states that all baggage must be box shaped with a sum of length, width, and height not exceeding 162 in. what are the dimensions and volume of a square-based box with the greatest volume under these conditions

The box will have all of the length, width, and height equal to 54 inches. The volume of the box will be 157464 cubic inches, or 91.125 cu feet. Since this problem says "square-based box", I'm assuming that the length and width of the box will be the same and the height may be whatever is needed to make the box fit. With that in mind, let's create a function based upon height that will calculate the volume of the box. Volume of box is V = L * W * H Since we have L and W the same, and the sun of L,W, and H can't exceed 162, we get: 162 = L + W + H 162 = 2L + H 162 - H = 2L (162 - H)/2 = L So the new expression for volume becomes V = L * W * H V = L^2 * H V = ((162 - H)/2)^2 * H V = (6561 - 81H + H^2/4) * H V = 6561H - 81H^2 + H^3/4 V = 0.25H^3 - 81H^2 + 6561H Since we're looking for a maximum, that will happen when the slope of the above equation is 0. And the first derivative will give us that slope. So calculate the first derivative, giving: V = 0.25H^3 - 81H^2 + 6561H V' = 0.75H^2 - 162H + 6561 And we now have a quadratic equation. Find the roots using the quadratic formula. They are: 54 and 162. We can immediately eliminate 162 as a possible solution since that would be a height of 162 which means that the length and width would both be 0 so the volume would be 0. But the value of 54 looks good. So H=54 Length and width will then be (162-54)/2 = 108/2 = 54 So all three dimensions of the largest volume box will be 54 inches. And the volume of the box will be 54^3 = 157464 in^3 or 157464/12^3 = 91.125 ft^3.

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It’s 24.6 after you round to the nearest tenths the answer should be 25.0