Question

suppose an airline policy states that all baggage must be box - shaped with a sum of length, width, and height not exceeding 138 in. what are the dimensions and volume of a square - based box with the greatest volume under these conditions?
the length of the square - end edge is
the box height is
(simplify your answers.)

Explanation:

Step1: Define variables

Let the length and width of the square - based box be \(x\) (since it's square - based, length = width), and the height be \(h\). The constraint is \(2x + h=138\), so \(h = 138 - 2x\). The volume formula for a box is \(V=x\times x\times h=x^{2}h\). Substitute \(h\) into the volume formula: \(V(x)=x^{2}(138 - 2x)=138x^{2}-2x^{3}\).

Step2: Find the derivative of the volume function

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the derivative \(V^\prime(x)\) of \(V(x)=138x^{2}-2x^{3}\) is \(V^\prime(x)=276x-6x^{2}\).

Step3: Set the derivative equal to zero

Set \(V^\prime(x) = 0\), so \(276x-6x^{2}=0\). Factor out \(6x\): \(6x(46 - x)=0\). This gives two solutions \(x = 0\) (which is not practical for a box) and \(x = 46\).

Step4: Use the second - derivative test

Find the second - derivative \(V^{\prime\prime}(x)\). Differentiate \(V^\prime(x)=276x - 6x^{2}\) with respect to \(x\), \(V^{\prime\prime}(x)=276-12x\). When \(x = 46\), \(V^{\prime\prime}(46)=276-12\times46=276 - 552=- 276<0\). So the volume is maximized at \(x = 46\).

Step5: Find the height

Substitute \(x = 46\) into the height formula \(h = 138-2x\). Then \(h=138 - 2\times46=138 - 92 = 46\).

Answer:

The length of the square - end edge is \(46\) in.
The box height is \(46\) in.