a box (with no top) will be made by cutting squares of equal size out of the corners of a 28 inch by 52 inch…

Explanation:

Step1: Define the variables

Let the side - length of the square cut from each corner be $x$ inches. Then the length of the box is $l = 52 - 2x$ inches, the width of the box is $w=28 - 2x$ inches, and the height of the box is $h = x$ inches. The volume $V$ of the box is given by the function $V(x)=(52 - 2x)(28 - 2x)x$.

Step2: Expand the function

[ \begin{align*} V(x)&=(52 - 2x)(28 - 2x)x\ &=(1456-56x - 104x+4x^{2})x\ &=(1456 - 160x+4x^{2})x\ &=1456x-160x^{2}+4x^{3} \end{align*} ]

Step3: Find the derivative

Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $V^\prime(x)=1456-320x + 12x^{2}$.

Step4: Set the derivative equal to zero

$12x^{2}-320x + 1456 = 0$. Divide through by 4 to simplify: $3x^{2}-80x + 364 = 0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$, here $a = 3$, $b=-80$, and $c = 364$. [ \begin{align*} x&=\frac{80\pm\sqrt{(-80)^{2}-4\times3\times364}}{2\times3}\ &=\frac{80\pm\sqrt{6400-4368}}{6}\ &=\frac{80\pm\sqrt{2032}}{6}\ &=\frac{80\pm4\sqrt{127}}{6}\ &=\frac{40\pm2\sqrt{127}}{3} \end{align*} ] $x_1=\frac{40 + 2\sqrt{127}}{3}\approx\frac{40+2\times11.27}{3}=\frac{40 + 22.54}{3}\approx20.85$ and $x_2=\frac{40 - 2\sqrt{127}}{3}\approx\frac{40-22.54}{3}\approx5.82$. But $x$ must satisfy $0\lt x\lt14$ (since $28-2x\gt0$). So we take $x=\frac{40 - 2\sqrt{127}}{3}\approx5.82$.

Step5: Calculate the maximum volume

Substitute $x\approx5.82$ into the volume function $V(x)=(52 - 2x)(28 - 2x)x$. $l = 52-2\times5.82=52 - 11.64 = 40.36$ inches, $w=28 - 2\times5.82=28 - 11.64 = 16.36$ inches, $h = 5.82$ inches. $V=(40.36)\times(16.36)\times5.82\approx40\times16\times6 = 3840$ cubic inches. [ \begin{align*} V&=(52-2\times5.82)(28 - 2\times5.82)\times5.82\ &=(52 - 11.64)(28 - 11.64)\times5.82\ &=40.36\times16.36\times5.82\ &\approx40.36\times16.36\times5.82\ &\approx3845 \end{align*} ]

Answer:

$3845$