a. squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring…

a. squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring 43 ft by 23 ft. the resulting piece of cardboard is then folded into a box without a lid. find the volume of the largest box that can be formed in this way.\nb. suppose that in part (a) the original piece of cardboard is a square with sides of length s. find the volume of the largest box that can be formed in this way.\na. the maximum volume of the box is approximately □ ft³ (round to the nearest hundredth as needed.)

a. squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring 43 ft by 23 ft. the resulting piece of cardboard is then folded into a box without a lid. find the volume of the largest box that can be formed in this way.\nb. suppose that in part (a) the original piece of cardboard is a square with sides of length s. find the volume of the largest box that can be formed in this way.\na. the maximum volume of the box is approximately □ ft³ (round to the nearest hundredth as needed.)

Answer

Explanation:

Step1: Find volume formula for part (a)

The length of the box after cutting and folding is $l = 43 - 2x$, the width is $w=23 - 2x$ and the height is $h = x$. So the volume $V(x)=(43 - 2x)(23 - 2x)x=4x^{3}-132x^{2}+989x$.

Step2: Differentiate the volume function

Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $V^\prime(x)=12x^{2}-264x + 989$.

Step3: Set the derivative equal to zero

$12x^{2}-264x + 989 = 0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$, here $a = 12$, $b=-264$, $c = 989$. So $x=\frac{264\pm\sqrt{(-264)^{2}-4\times12\times989}}{2\times12}=\frac{264\pm\sqrt{69696-47472}}{24}=\frac{264\pm\sqrt{22224}}{24}=\frac{264\pm149.08}{24}$. We get two solutions $x_1=\frac{264 + 149.08}{24}\approx17.21$ and $x_2=\frac{264-149.08}{24}\approx4.79$. But $x_1$ is not valid since $2x_1=34.42>23$. So we use $x = 4.79$.

Step4: Calculate the maximum volume

Substitute $x = 4.79$ into $V(x)$: $V(4.79)=(43-2\times4.79)(23 - 2\times4.79)\times4.79=(43 - 9.58)(23-9.58)\times4.79=33.42\times13.42\times4.79\approx2148.05$.

Answer:

$2148.05$