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Question 1 of 5

An open rectangular box is to be made by cutting out square corners from a square

$90xx90$ cm piece of cardboard and folding up the sides. What is the maximum volume of the box?

$V=$ (54000) $\text(cm)^3$

Question 2 of 5

$24$ cm piece of wire is bent into the shape of a rectangle. Find the maximum area of this rectangle.

$A=$ (36) $\text(cm)^2$

Question 3 of 5

3. Question
Maximize the cube below given that the sum of its dimensions is $60\text(cm)$
- (8000) $\text(cm)^3$
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Volume of a Cube

$V=lwh$

Form the first equation using the given information

$x+x+y$ $=$ $60$ Sum of dimensions equal $60\text(cm)$

$2x+y$ $=$ $60$ Equation $1$

Form another equation using the Volume Formula

$V$ $=$ $lwh$ Volume Formula

$V$ $=$ $x*x*y$ Substitute given dimensions

$V$ $=$ $x^2y$ Equation $2$

Since we are hoping to minimize the Volume, choose Equation $2$ as the main function

Write Equation $1$ in terms of $y$ and use it so the main equation only has one variable

$2x+y$ $=$ $60$ Equation $1$

$y$ $=$ $60-2x$

Substitute this to Equation $2$

$V$ $=$ $x^2y$ Equation $2$

$V$ $=$ $x^2(60-2x)$ Substitute $y$ from previous step

$V$ $=$ $60x^2-2x^3$

Start optimizing the function by getting its first derivative and equating it to $0$ to solve for $x$

$V$ $=$ $60x^2-2x^3$

$V’$ $=$ $120x-6x^2$

$120x-6x^2$ $=$ $0$ Equate $V’$ to $0$

$6x(20-x)$ $=$ $0$ Factor out $6x$

$x$ $=$ $0,20$

Since $x$ is a length, it cannot be $0$ . Hence, we choose $x=20$

Substitute this value of $x$ to the second derivative of $V$

$V’$ $=$ $120x-6x^2$

$V”$ $=$ $120-12x$

$=$ $120-12($ $20$ $)$ Substitute $x=20$

$=$ $120-240$

$=$ $-120$

This value is negative, which means $x=20$ yields the maximum value

Compute for $y$ by substituting $x$ to Equation $1$

$2x+y$ $=$ $60$ Equation $1$

$2(20)+y$ $=$ $60$ Substitute $x=20$

$40+y$ $=$ $60$

$y$ $=$ $60-40$

$y$ $=$ $20$

Finally, compute for the maximum volume by substituting $x$ and $y$ to $V$

$x=20$

$y=20$

$V$ $=$ $x^2y$

$=$ $(20)^2(20)$ Substitute $x$ and $y$

$=$ $20^3$

$=$ $8000\text(cm)^3$

$8000\text(cm)^3$

Question 4 of 5

A company that manufactures soft drinks wants to save money by restricting each can to have a Surface Area of

$726pi\text(cm)^2$ . Find the maximum volume of the can with this restriction.

1. $14641pi\text(cm)^3$
2. $1221pi\text(cm)^3$
3. $3993pi\text(cm)^3$
4. $2662pi\text(cm)^3$

Question 5 of 5

A page is printed with a margin, as shown below. Given that the page area is

$288\text(cm)^2$ , maximise the dimensions so that the printed internal area is maximised.

$x=$ (12)

$y=$ (24)

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Perimeter of a Rectangle

$P=2x+2y$

Area of a Rectangle

$A=xy$

Form the first equation using the Area Formula for the Page Area

$A$	$=$	$xy$	Area Formula
$288$	$=$	$xy$
$xy$	$=$	$288$	Equation $1$

Form another equation using the Area Formula for the Printed Area

$A$	$=$	$xy$	Area Formula
$A_p$	$=$	$(x-1-1)(y-2-2)$	Substitute given dimensions
$A_p$	$=$	$(x-2)(y-4)$	Equation $2$

Since we are hoping to maximize the Printed Area, choose Equation

$2$ as the main function

Write Equation

$1$ in terms of

$y$ and use it so the main equation only has one variable

$xy$	$=$	$288$	Equation $1$
$xy$ $-:x$	$=$	$288$ $-:x$	Divide both sides by $x$

$y$	$=$	$288/x$	Updated Equation $1$

Substitute this to Equation

$2$

$A_p$	$=$	$(x-2)(y-4)$	Equation $2$

	$=$	$(x-2)(288/x-4)$	Substitute $y$ from previous step

	$=$	$x(288/x)-4x-2(288/x)+8$

	$=$	$288-4x-576/x+8$

	$=$	$296-576/x-4x$

$A_p$	$=$	$296-576x^(-1)-4x$	Updated Equation $2$

Start optimizing the function by getting its first derivative and equating it to

$0$ to solve for

$x$

$A_p$	$=$	$296-576x^(-1)-4x$
$A’_p$	$=$	$-(-576x^(-2))-4$

$A’_p$	$=$	$576/(x^2)-4$

$576/(x^2)-4$	$=$	$0$	Equate $A’_p$ to $0$

$576/(x^2)$	$=$	$4$

$576/(x^2)$ $times x^2$	$=$	$4$ $times x^2$	Multiply both sides by $x^2$

$576$	$=$	$4x^2$
$576$ $-:4$	$=$	$4x^2$ $-:4$	Divide both sides by $4$
$144$	$=$	$x^2$
$sqrt144$	$=$	$sqrt(x^2)$	Get the square root of both sides
$12$	$=$	$x$
$x$	$=$	$12$

Substitute this value of

$x$ to the second derivative of

$A_p$

$A’_p$	$=$	$576x^(-2)-4$
$A”_p$	$=$	$-1152x^(-3)$

	$=$	$-1152/(x^3)$

	$=$	$-\frac{1152}{\color{#007DDC}{12}^3}$	Substitute $x=12$

	$=$	$-1152/1728$

This value is negative, which means $x=12$ yields the maximum value

Compute for

$y$ by substituting

$x$ to the Updated Equation

$1$

$y$	$=$	$288/x$	Updated Equation $1$

$y$	$=$	$288/12$	Substitute $x=12$

$y$	$=$	$24$

$x=12$

$y=24$

$P$	$=$	$2l+2w$	Perimeter Formula
$24$	$=$	$2x+2y$	Substitute values
$24$ $-:2$	$=$	$(2x+2y)$ $-:2$	Divide both sides by $2$
$12$	$=$	$x+y$
$12-x$	$=$	$y$
$y$	$=$	$12-x$	Leave only $y$ on the left side

$A’$	$=$	$12-2x$
$A”$	$=$	$-2$

$SA$	$=$	$2pir^2+2pirh$	Surface Area Formula
$726pi$	$=$	$2pir^2+2pirh$	Substitute given value
$726pi-2pir^2$	$=$	$2pirh$
$2pirh$	$=$	$726pi-2pir^2$
$2pirh$ $-:2pi$	$=$	$(726pi-2pir^2)$ $-:2pi$	Divide both sides by $2pi$
$rh$	$=$	$363-r^2$
$rh$ $-:r$	$=$	$(363-r^2)$ $-:r$	Divide both sides by $r$

$h$	$=$	$(363-r^2)/r$

$V$	$=$	$363pir-pir^3$
$V’$	$=$	$363pi-3pir^2$
$363pi-3pir^2$	$=$	$0$	Equate $V’$ to $0$
$363pi$	$=$	$3pir^2$
$363pi$ $-:pi$	$=$	$3pir^2$ $-:pi$	Divide both sides by $pi$
$363$	$=$	$3r^2$
$363$ $-:3$	$=$	$3r^2$ $-:3$	Divide both sides by $3$
$121$	$=$	$r^2$
$sqrt121$	$=$	$sqrt(r^2)$	Get the square root of both sides
$11$	$=$	$r$
$r$	$=$	$11$

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Quiz summary

Information

1. Question

Hint

Volume of a Rectangular Prism

2. Question

Hint

Perimeter of a Rectangle

Area of a Rectangle

3. Question

Hint

Volume of a Cube

4. Question

Hint

Volume of a Cylinder

Surface Area of a Cylinder

5. Question

Hint

Perimeter of a Rectangle

Area of a Rectangle

Quizzes

$V$	$=$	$lxxwxxh$	Volume Formula
	$=$	$(90-2x)(90-2x)x$	Substitute values
	$=$	$(8100-180x-180x+4x^2)x$
	$=$	$(8100-360x+4x^2)x$
$V$	$=$	$4x^3-360x^2+8100x$	Arrange the terms in descending powers

$V$	$=$	$4x^3-360x^2+8100x$
$V’$	$=$	$12x^2-720x+8100$
	$=$	$12(x^2-60x+675)$
	$=$	$12(x-45)(x-15)$	$=$	$0$
		$x=$ $45,15$

$V’$	$=$	$12x^2-720x+8100$
$V”$	$=$	$24x-720$
	$=$	$24($ $45$ $)-720$
	$=$	$1080-720$
	$=$	$360$

$V$	$=$	$4x^3-360x^2+8100x$
	$=$	$4(15^3)-360(15^2)+8100(15)$	Substitute $x=15$
	$=$	$4(3375)-360(225)+8100(15)$
	$=$	$13500-81000+121500$
	$=$	$54 000\text(cm)^3$

$A$	$=$	$lw$	Area Formula
$A$	$=$	$xy$	Substitute $x$ and $y$
$A$	$=$	$x(12-x)$	Substitute $y=12-x$
$A$	$=$	$12x-x^2$

$12$	$=$	$2x$
$12$ $-:2$	$=$	$2x$ $-:2$	Divide both sides by $2$
$6$	$=$	$x$
$x$	$=$	$6$

$A$	$=$	$12x-x^2$
	$=$	$12(6)-(6)^2$	Substitute $x=6$
	$=$	$72-36$
	$=$	$36\text(cm)^2$

$x+x+y$	$=$	$60$	Sum of dimensions equal $60\text(cm)$
$2x+y$	$=$	$60$	Equation $1$

$V$	$=$	$lwh$	Volume Formula
$V$	$=$	$xxy$	Substitute given dimensions
$V$	$=$	$x^2y$	Equation $2$

$V$	$=$	$x^2y$	Equation $2$

$V$	$=$	$x^2(60-2x)$	Substitute $y$ from previous step

$V$	$=$	$60x^2-2x^3$

$V’$	$=$	$120x-6x^2$
$V”$	$=$	$120-12x$
	$=$	$120-12($ $20$ $)$	Substitute $x=20$
	$=$	$120-240$
	$=$	$-120$

$V$	$=$	$x^2y$
	$=$	$(20)^2(20)$	Substitute $x$ and $y$
	$=$	$20^3$
	$=$	$8000\text(cm)^3$

$V$	$=$	$pir^2h$	Volume Formula
	$=$	$pir^2((363-r^2)/r)$	Substitute $h$ from previous step
	$=$	$pir(363-r^2)$
$V$	$=$	$363pir-pir^3$

$V$	$=$	$363pir-pir^3$
	$=$	$363pi(11)-pi(11^3)$	Substitute $r$
	$=$	$3993pi-1331pi$
	$=$	$2662pi\text(cm)^3$