Answer to Question #205437 in Calculus for yummy

Question

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.

Answer 1

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

$$V=length\times width \times height$$

$$V=(20-2x)(16-2x)(x)$$

$$=320x-40x^2-32x^2+4x^3$$

$$=4x^3-72x^2+320x$$

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)

$$V'(x)=0=>12x^2-144x+320=0$$

$$3x^2-36x+80=0$$

$D=(-36)^2-4(3)(80)=336$

$$x=\dfrac{36\pm\sqrt{336}}{2(3)}=6\pm\dfrac{2\sqrt{21}}{3}$$

$$x_1=6-\dfrac{2\sqrt{21}}{3}, 0<x_1<8$$

$$x_2=6-\dfrac{2\sqrt{21}}{3}, x_2>8$$

Critical numbers: $6-\dfrac{2\sqrt{21}}{3}, 6+\dfrac{2\sqrt{21}}{3}.$

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

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

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

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

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

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

iii) Length of the base of the box

$$20-2(6-\dfrac{2\sqrt{21}}{3})=(8+\dfrac{4\sqrt{21}}{3})\ m\approx14.110\ m$$

Width of the base of the box

$$16-2(6-\dfrac{2\sqrt{21}}{3})=(4+\dfrac{4\sqrt{21}}{3})\ m\approx10.110\ m$$

Height of the box

$$(6-\dfrac{2\sqrt{21}}{3})\ m\approx2.945\ m$$

$$V\approx420.011\ m^3$$

2) Let $2x=$ the width of the rectangle

Let $y=$ the height of the rectangle.

Then

$$Perimeter=2y+x+\dfrac{1}{2}(2\pi (\dfrac{1}{2}x))=4$$

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 $x=\dfrac{8}{4+\pi}.$

$$y=2-(\dfrac{2+\pi}{4})(\dfrac{8}{4+\pi})$$

$$=\dfrac{8+2\pi-4-2\pi}{4+\pi}=\dfrac{4}{4+\pi}$$

$x=\dfrac{8}{4+\pi}\ m\approx1.120\ m$

$y=\dfrac{4}{4+\pi}\ m\approx0.560\ m$

(iii)

$$A_{max}=2(\dfrac{8}{4+\pi})-(\dfrac{4+\pi}{8})(\dfrac{8}{4+\pi})^2$$

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

$A_{max}=1.1202\ m^2$