Naij Observers

Exercise

The sum of two positive numbers is 20

. One of the numbers is multiplied by the square of the other. Find the numbers that make this product a maximum.

Let the first number be x

and the second number be y and let the product be P. We get the following two equations:

x+yxy2=20=P

Rearranging the first equation and substituting into the second gives:

P=(20−x)2x=400x−40x2+x3

Differentiating and setting to 0
gives:

P′0=400−80x+3x2=3x2−80x+400=(3x−20)(x−20)

Therefore, x=20
or x=203.
If x=20
then y=0 and the product is a minimum, not a maximum.
Therefore, x=203
and y=20−203=403.
Therefore the two numbers are 203
and 403 (approximating to the nearest integer gives 7 and 13

).

A wooden block is made as shown in the diagram. The ends are right-angled triangles having sides 3x

, 4x and 5x. The length of the block is y. The total surface area of the block is 3 600 cm2

.

Show that y=300−x2x

.

We start by finding the surface area of the prism:

Surface area3 600=2(12b×h)+3xy+4xy+5xy=(3x×4x)+12xy=12x2+12xy

Solving for y
gives:

12xyyy=3 600−12x2=3 600−12x212x=300−x2x

Find the value of x

for which the block will have a maximum volume.
(Volume = area of base ×

height)

Start by finding an expression for volume in terms of x

:

VV= area of triangle ×y=6x2×300−x2x=6x(300−x2)=1 800x−6x3

Now take the derivative and set it equal to 0
:

V′018x2x2x=1 800−18x2=1 800−18x2=1 800=100=±10

Since the length can only be positive, x=10
∴x=10 cm

Determine the shortest vertical distance between the curves of f

and g if it is given that:

f(x)and g(x)=−x2+2x+3=8x,x>0

Let the distance P(x)=g(x)−f(x)=8x−(−x2+2x+3)=8x+x2−2x−3

To minimise the distance between the curves, let P′(x)=0:

P′(x)0∴0000∴x=2∴x=−8x2+2x−2(x≠0)=−8x2+2x−2=−8+2x3−2x=2x3−2x−8=x3−x−4=(x−2)(x2+x+2) or x=−1±(1)2−4(1)(2)−−−−−−−−−−−√2(1)= no real solutions =2

Therefore, the shortest distance:

P(2)=8(2)+(2)2−2(2)−3=4+4−4−3=1 unit

The diagram shows the plan for a verandah which is to be built on the corner of a cottage. A railing ABCDE

is to be constructed around the four edges of the verandah.
If AB=DE=x
and BC=CD=y, and the length of the railing must be 30 m, find the values of x and y

for which the verandah will have a maximum area.

We need to determine an expression for the area in terms of only one variable.
The perimeter is:

P3015y=2x+2y=2x+2y=x+y=15−x

The area is:

A=y2−(y−x)2=y2−(y2−2xy+x2)=y2−y2+2xy−x2=2xy−x2

We use the expression for perimeter to eliminate the y
variable so that we have an expression for area in terms of x only:

A(x)=2x(15−x)−x2=30x−2x2−x2=30x−3x2

To find the maximum, we need to take the derivative and set it equal to 0
:

A′(x)06xx=30−6x=30−6x=30=5

Therefore, x=5 m
and substituting this value back into the formula for perimeter gives y=10 m

.

A rectangular juice container, made from cardboard, has a square base and holds 750 cm3

of juice. The container has a specially designed top that folds to close the container. The cardboard needed to fold the top of the container is twice the cardboard needed for the base, which only needs a single layer of cardboard.

If the length of the sides of the base is x

cm, show that the total area of the cardboard needed for one container is given by:

A(in square centimetres)=3 000x+3x2

V750∴hASubstitute hA=x2h=x2h=750x2= area of sides + area of base + area of top =4xh+x2+2x2=4xh+3x2=750x2:=4x(750x2)+3x2=3000x+3x2

Determine the dimensions of the container so that the area of the cardboard used is minimised.

A(x)A′(x)∴06xx3∴x∴h=3000x+3x2=−3000x2+6x=−3000x2+6x=3000x2=500=500−−−√3≈7,9 cm=750(7,9)2≈12,0 cm