Question

You are planning to make an open rectangular box from a 25​-in.-by-49​-in. piece of cardboard by cutting congruent squares from the corners and folding up the sides. What are the dimensions of the box of largest volume you can make this​ way, and what is its​ volume?

Answer 1

Answer:

The dimensions of the box are 38.48 x 14.48 x 5.26 in.

Step-by-step explanation:

We will have a piece of cardboard with squares of side x cut from the corners to make a open box.

The volume of the box can be written as:

$V=A\cdot B\cdot x\\\\V=(49-2x)\cdot (25-2x)\cdot x\\\\V=x\cdot (4x^2-2x*(49+25)+49*25)\\\\V=4x^3-148x^2+1225x$

To calculate the maximum volume, we derive and equal to zero

$dV/dx=12x^2-296x+1225=0$

We apply the quadratic equation to know the roots of the equation

$x=\frac{-b\pm\sqrt{b^24ac}}{2a}\\\\x=\frac{296\pm\sqrt{296^2-4*12*1225}}{2*12} =\frac{296\pm\sqrt{87616-58800}}{24}= \frac{296\pm\sqrt{28816}}{24}\\\\x=\frac{296\pm169.75}{24}= 12.333\pm7.073\\\\\\x_1=12.333+7.073=19.406\\\\x_2=12.333-7.073=5.26$

The first solution is physically impossible, as the side that is cut would be bigger than 25.

So the real solution is for x=5.26 in.

Then, the dimensions of the box are:

$A=49-2x=49-2*5.26=49-10.52=38.48\\\\B=25-2x=25-10.52=14.48$