# Answer in Calculus for yummy #205437

1) A designer of a box making company wants to produce an open box from a piece of card

with the width of 16cm and the length of 20cm. The box is made by cutting out squares of

length x at each corners.

i)Show that the volume of the box is V=4x^3- 72x^2-320x

ii) Find the value of x so that the volume of the box is maximum.

iii) Find the dimensions of the box so that the volume of the box is maximum.

2) A glass window frame is in the form of a rectangle at the bottom and a semicircle at the

top as shown in Figure 1, has a perimeter of 4 m.

i) Show that the area of the glass window is given by

A = 2x − (4 + π8 )x^2

ii) Find the values of x and y that will maximized the area of the glass window.

(0.56 m)

iii) Hence, find the maximum area of the glass window.

1) Let “x=” the side of the square

“0<x<16/2”

The length of the base of the box is “(20-2x)” cm, the width of the base of the box is “(16-2x)” cm, and the height of the box is “x” cm.

i) The volume “V” of the box will be

“V=lengthtimes width times height”

“V=(20-2x)(16-2x)(x)”

“=320x-40x^2-32x^2+4x^3”

“=4x^3-72x^2+320x”

ii) Find the first derivative of “V” with respect to “x”

“V'(x)=(4x^3-72x^2+320x)’=12x^2-144x+320”

Find the critical number(s)

“V'(x)=0=>12x^2-144x+320=0”

“3x^2-36x+80=0”

“D=(-36)^2-4(3)(80)=336”

“x=dfrac{36pmsqrt{336}}{2(3)}=6pmdfrac{2sqrt{21}}{3}”

“x_1=6-dfrac{2sqrt{21}}{3}, 0<x_1<8”

“x_2=6-dfrac{2sqrt{21}}{3}, x_2>8”

Critical numbers: “6-dfrac{2sqrt{21}}{3}, 6+dfrac{2sqrt{21}}{3}.”

Since “0<x<8,” we consider “6-dfrac{2sqrt{21}}{3}.”

If “0<x<6-dfrac{2sqrt{21}}{3},” “V'(x)>0, V” increases.

If “6-dfrac{2sqrt{21}}{3}<x<8, V'(x)<0,V” decreases.

The function “A” has a local maximum at “x=6-dfrac{2sqrt{21}}{3}.”

Since the function “V” has the only extremum on “(0, 8),” then the function “V” has the absolute maximum on “(0, 8)” at “x=6-dfrac{2sqrt{21}}{3}.”

The volume of the box is maximum when “x=6-dfrac{2sqrt{21}}{3}.”

iii) Length of the base of the box

“20-2(6-dfrac{2sqrt{21}}{3})=(8+dfrac{4sqrt{21}}{3}) mapprox14.110 m”

Width of the base of the box

“16-2(6-dfrac{2sqrt{21}}{3})=(4+dfrac{4sqrt{21}}{3}) mapprox10.110 m”

Height of the box

“(6-dfrac{2sqrt{21}}{3}) mapprox2.945 m”

“Vapprox420.011 m^3”

2) Let “2x=” the width of the rectangle

Let “y=” the height of the rectangle.

Then

“Perimeter=2y+x+dfrac{1}{2}(2pi (dfrac{1}{2}x))=4”

Solve for “y”

“y=2-(dfrac{2+pi}{4})x”

1) Area “A” of the window

“A=xy+dfrac{1}{2}pi(dfrac{x}{2})^2”

“=2x-(dfrac{2+pi}{4})x^2+dfrac{pi}{8}x^2”

“=2x-(dfrac{4+pi}{8})x^2”

ii) Find the first derivative of “A” with respect to “x”

“A'(x)=(2x-(dfrac{4+pi}{8})x^2)’=2-(dfrac{4+pi}{4})x”

Find the critical number(s)

“A'(x)=0=>2-(dfrac{4+pi}{4})x=0”

“x=dfrac{8}{4+pi}”

If “0<x<dfrac{8}{4+pi}, A'(x)>0, A” increases.

If “dfrac{8}{4+pi}<x<4, A'(x)<0, A” decreases.

The function “A” has a local maximum at “x=dfrac{8}{4+pi}.”

Since the function “A” has the only extremum on “(0, 4),” then the function “A” has the absolute maximum on “(0, 4)” at “x=dfrac{8}{4+pi}.”

“y=2-(dfrac{2+pi}{4})(dfrac{8}{4+pi})”

“=dfrac{8+2pi-4-2pi}{4+pi}=dfrac{4}{4+pi}”

“x=dfrac{8}{4+pi} mapprox1.120 m”

“y=dfrac{4}{4+pi} mapprox0.560 m”

(iii)

“A_{max}=2(dfrac{8}{4+pi})-(dfrac{4+pi}{8})(dfrac{8}{4+pi})^2”

“=dfrac{16-8}{4+pi}=dfrac{8}{4+pi}(m^2)approx1.1202( m^2)”

“A_{max}=1.1202 m^2”