Question

a candy box is made from a piece of cardboard that measures 39 by 21 inches. squares of equal size will be cut out of each corner. the sides will then be folded up to form a rectangular box. what size square should be cut from each corner to obtain maximum volume?
a square with a side of length □ inches should be cut away from each corner to obtain the maximum volume.
(round to the nearest hundredth as needed.)

Explanation:

Step1: Set up the volume function

Let the side - length of the square cut from each corner be \(x\) inches.
The length of the box is \(L=(39 - 2x)\) inches, the width of the box is \(W=(21 - 2x)\) inches, and the height of the box is \(H = x\) inches.
The volume \(V\) of the box is given by the formula \(V=(39 - 2x)(21 - 2x)x\).
Expand the function:

$$
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$$

Step2: Find the derivative of the volume function

Differentiate \(V(x)=4x^{3}-120x^{2}+819x\) with respect to \(x\).
Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), we have \(V^\prime(x)=12x^{2}-240x + 819\).

Step3: Set the derivative equal to zero

Set \(V^\prime(x)=0\), so \(12x^{2}-240x + 819 = 0\).
Divide through by \(3\): \(4x^{2}-80x + 273=0\).
Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 4\), \(b=-80\), and \(c = 273\).

$$
LATEXBLOCK1
$$

We get \(x_1=\frac{80 + 45.08}{8}=\frac{125.08}{8}=15.635\) and \(x_2=\frac{80 - 45.08}{8}=\frac{34.92}{8}=4.365\).

Step4: Check the domain and the second - derivative

The domain of \(x\) is \(0\lt x\lt10.5\) (since \(21-2x>0\)). So \(x = 15.635\) is not in the domain.
Differentiate \(V^\prime(x)\) to get the second - derivative \(V^{\prime\prime}(x)=24x-240\).
When \(x = 4.365\), \(V^{\prime\prime}(4.365)=24\times4.365-240=104.76-240=-135.24<0\).

Answer:

\(4.37\) inches