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 35 ft by 19 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. b. 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. a. 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 35 ft by 19 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. b. 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. a. the maximum volume of the box is approximately □ ft³. (round to the nearest hundredth as needed.)

Answer

Explanation:

Step1: Find the dimensions of the box

The length of the box after cutting squares of side - length $x$ from a $35$ ft by $19$ ft cardboard and folding is $l = 35 - 2x$, the width is $w=19 - 2x$, and the height is $h = x$. The volume function $V(x)=(35 - 2x)(19 - 2x)x=x(665-70x - 38x+4x^{2})=4x^{3}-108x^{2}+665x$.

Step2: Find the derivative of the volume function

Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $V^\prime(x)=12x^{2}-216x + 665$.

Step3: Set the derivative equal to zero and solve for $x$

We use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $12x^{2}-216x + 665 = 0$, where $a = 12$, $b=-216$, and $c = 665$. [x=\frac{216\pm\sqrt{(-216)^{2}-4\times12\times665}}{2\times12}=\frac{216\pm\sqrt{46656-31920}}{24}=\frac{216\pm\sqrt{14736}}{24}=\frac{216\pm121.4}{24}] We get $x_1=\frac{216 + 121.4}{24}\approx14.06$ and $x_2=\frac{216-121.4}{24}\approx3.94$. But $x$ must satisfy $0\lt x\lt\frac{19}{2}=9.5$. So we take $x\approx3.94$.

Step4: Find the maximum volume

Substitute $x\approx3.94$ into the volume function $V(x)$. [V(3.94)=(35-2\times3.94)(19 - 2\times3.94)\times3.94=(35 - 7.88)(19 - 7.88)\times3.94=27.12\times11.12\times3.94] [V(3.94)\approx27.12\times11.12\times3.94 = 27.12\times43.8 = 1187.86]

Answer:

$1187.86$