A manufacturer cuts squares from the corners of a rectangular piece of sheet met

Question

A manufacturer cuts squares from the corners of a rectangular piece of sheet metal that measures 3 inches by 8 inches (see Figure 1). The manufacturer then folds the metal upward to make an open-topped box (see Figure 2). Letting x represent the side-lengths (in inches) of the squares, use the ALEKS graphing calculator to find the value of x that maximizes the volume enclosed by this box. Then give the maximum volume. Round your responses to two decimal places Figure 1 Figure 2 value of x that maximizes volume: in Maximum volume in

Explanation / Answer

after the square of side x being cut the length of box become = 8-2x inch

width = 3-2x inch

height = x inch

So the volume of box V(x) = length * width * height

=(8-2x)(3-2x)x = 24x-22x^2+4x^3

V'(x) = 24-44x+12x^2

V''(x)= -44+24x

WE maximize volume V(x) with respect to x

for critical point solve V'(x)=0 for x as,

24-44x+12x^2=0

4(6-11x+3x^2)=0

4(3-x)(2-3x)=0

x= 2/3, 3

As 3-2x= 3-2*3= -3<0 so width is negative for x=3 which is not possible so discard this value

Note V''(2/3) = -28<0 so by second derivative test x=2/3 ~ 0.67 is a point of maxima

Thus value of x that maximize volume is = 0.67 inch

maximum volume V(0.67) = 7.41 inch^3

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