MATH SOLVE

9 months ago

Q:
# A candy box is made from a piece of cardboard that measures 25 by 14 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 maximumâ volume?

Accepted Solution

A:

Let the size of the squares to be cut out be x, then after the squares are cut out from each corner, the length of the rectangular box to be formed becomes 25 - 2x, the width 14 - 2x and the height x,

The volume of a rectangular box is given by length x width x height

[tex]=(25-2x)(14-2x)x=x(350-78x+4x^2) \\ \\ =350x-78x^2+4x^3[/tex]

For maximum volume,

[tex] \frac{dV}{dx} =0 \\ \\ \Rightarrow350-156x+12x^2=0 \\ \\ \Rightarrow x= \frac{-(-156)\pm\sqrt{(-156)^2-4(12)(350)}}{2(12)} \\ \\ = \frac{156\pm\sqrt{24,336-16,800}}{24} = \frac{156\pm\sqrt{7,536}}{24} \\ \\ = \frac{156\pm86.81}{24} = \frac{156+86.81}{24} \ or \ \frac{156-86.81}{24} \\ \\ = \frac{242.81}{24} \ or \ \frac{69.19}{24} = 10.12\ or \ 2.88[/tex]

But the size of the square cut out cannot be 10.12, for 10.12 inches cannot be cut out of both sides of the width, since the width is 14 inches.

Therefore, the size of the square to be cut out for maximum volume is 2.88 inches.

The volume of a rectangular box is given by length x width x height

[tex]=(25-2x)(14-2x)x=x(350-78x+4x^2) \\ \\ =350x-78x^2+4x^3[/tex]

For maximum volume,

[tex] \frac{dV}{dx} =0 \\ \\ \Rightarrow350-156x+12x^2=0 \\ \\ \Rightarrow x= \frac{-(-156)\pm\sqrt{(-156)^2-4(12)(350)}}{2(12)} \\ \\ = \frac{156\pm\sqrt{24,336-16,800}}{24} = \frac{156\pm\sqrt{7,536}}{24} \\ \\ = \frac{156\pm86.81}{24} = \frac{156+86.81}{24} \ or \ \frac{156-86.81}{24} \\ \\ = \frac{242.81}{24} \ or \ \frac{69.19}{24} = 10.12\ or \ 2.88[/tex]

But the size of the square cut out cannot be 10.12, for 10.12 inches cannot be cut out of both sides of the width, since the width is 14 inches.

Therefore, the size of the square to be cut out for maximum volume is 2.88 inches.