Maximisation/Minimisation Problems (1 Viewer)

acevipa · Mar 12, 2009

I'm having trouble with these few questions:

1. The diagonal of the base of a box in the shape of a cuboid has a length of 10cm. If one edge of the base has a length of x cm, express, in terms of x, the length of the other edge of the base. If the height of the box is equal to the length of this other edge express the volume of the box in terms of x and find the maximum volume of the box.

2. A piece of wire of length 30cm is cut into two sections. Each section is then bent into the shape of a square. Find the smallest possible value of the sum of the areas of the two squares.

flon · Mar 12, 2009

Question 2.

let x and y be the two sections.

x + y = 30
y = 30 - x -- > 1.

A = (x/4)^2 + (y/4)^2

Sub 1. into A.

A = (x/4)^2 + ((30-x)/4)^2

Differentiate A. To find the minimum value for X. then sub back into x+y = 30 to find y value. Once found both values. sub into A. to find smallest possible value.

kurt.physics · Mar 12, 2009

acevipa said:
I'm having trouble with these few questions:

1. The diagonal of the base of a box in the shape of a cuboid has a length of 10cm. If one edge of the base has a length of x cm, express, in terms of x, the length of the other edge of the base. If the height of the box is equal to the length of this other edge express the volume of the box in terms of x and find the maximum volume of the box.

I do believe the shape in question we are looking at is a rectangular prism. The diagonal of the base of the prism is 10, one side is x. Let the other side be y; then:

$x^2+y^2 = 10^2 \,\,\,\,$(from pythagoras)$$

$\therefore y = \sqrt{100-x^2} \,\,.\,.\,. (1)$

$Now,$

$$Volume$\,\,= x\times y\times y$

$V = x y^2\,\,.\,.\,.\,(2)$

$substituting$\,\,y = \sqrt{100-x^2}\,\,$into$\,\,(2)\,\,$gives,$

$V = x (\sqrt{100-x^2})^2$

$V = x (100-x^2)$

$V = 100x -x^3$

$\frac{dV}{dx} = 100 -3x^2$

$$For stationary points,$\,\, \frac{dV}{dx} = 0$

$$ie$\,\, 100 - 3x^2 = 0$

$3x^2 = 100$

$x^2 = \frac{100}{3}$

$x = \pm\frac{10}{\sqrt{3} }$

$x = \pm\frac{10\sqrt{3}}{3}$

$As we know that x can only be positive (as it is a length), then we only need to consider:$

$x = \frac{10\sqrt{3}}{3}$

$Now,$

$\frac{d^2V}{dx^2} = -6x$

$$when$\,\,x=\frac{10\sqrt{3}}{3},\,\,\frac{d^2V}{dx^2} < 0$

$\therefore x=\frac{10\sqrt{3}}{3}\,\, $is a maximum$$

$To find the maximum volume, we substitute this back into the equation$

$V = x (100-x^2)$

$= \frac{10\sqrt{3}}{3} (100- ( \frac{10\sqrt{3}}{3} )^2 )$

$= \frac{2000\sqrt{3} }{9} cm^3$

$\approx 384.9 cm^3$

Official · Mar 12, 2009

kurt.physics said:
I do believe the shape in question we are looking at is a rectangular prism. The diagonal of the base of the prism is 10, one side is x. Let the other side be y; then:

$x^2+y^2 = 10^2 \,\,\,\,$(from pythagoras)$$

$\therefore y = \sqrt{100-x^2} \,\,.\,.\,. (1)$

$Now,$

$$Volume$\,\,= x\times y\times y$

$V = x y^2\,\,.\,.\,.\,(2)$

$substituting$\,\,y = \sqrt{100-x^2}\,\,$into$\,\,(2)\,\,$gives,$

$V = x (\sqrt{100-x^2})^2$

$V = x (100-x^2)$

$V = 100x -x^3$

$\frac{dV}{dx} = 100 -3x^2$

$$For stationary points,$\,\, \frac{dV}{dx} = 0$

$$ie$\,\, 100 - 3x^2 = 0$

$3x^2 = 100$

$x^2 = \frac{100}{3}$

$x = \pm\frac{10}{\sqrt{3} }$

$x = \pm\frac{10\sqrt{3}}{3}$

$As we know that x can only be positive (as it is a length), then we only need to consider:$

$x = \frac{10\sqrt{3}}{3}$

$Now,$

$\frac{d^2V}{dx^2} = -6x$

$$when$\,\,x=\frac{10\sqrt{3}}{3},\,\,\frac{d^2V}{dx^2} < 0$

$\therefore x=\frac{10\sqrt{3}}{3}\,\, $is a maximum$$

$To find the maximum volume, we substitute this back into the equation$

$V = x (100-x^2)$

$= \frac{10\sqrt{3}}{3} (100- ( \frac{10\sqrt{3}}{3} )^2 )$

$= \frac{2000\sqrt{3} }{9} cm^3$

$\approx 384.9 cm^3$

Ah nice explanation! Repped.

acevipa · Mar 12, 2009

Thanks. Well explained.

Aplus · Mar 13, 2009

kurt.physics said:
$x^2+y^2 = 10^2 \,\,\,\,$(from pythagoras)$$

$\therefore y = \sqrt{100-x^2} \,\,.\,.\,. (1)$

$Now,$

$$Volume$\,\,= x\times y\times y$

$V = x y^2\,\,.\,.\,.\,(2)$

$substituting$\,\,y = \sqrt{100-x^2}\,\,$into$\,\,(2)\,\,$gives,$

$V = x (\sqrt{100-x^2})^2$

$V = x (100-x^2)$

$V = 100x -x^3$

$\frac{dV}{dx} = 100 -3x^2$

$$For stationary points,$\,\, \frac{dV}{dx} = 0$

$$ie$\,\, 100 - 3x^2 = 0$

$3x^2 = 100$

$x^2 = \frac{100}{3}$

$x = \pm\frac{10}{\sqrt{3} }$

$x = \pm\frac{10\sqrt{3}}{3}$

$As we know that x can only be positive (as it is a length), then we only need to consider:$

$x = \frac{10\sqrt{3}}{3}$

$Now,$

$\frac{d^2V}{dx^2} = -6x$

$$when$\,\,x=\frac{10\sqrt{3}}{3},\,\,\frac{d^2V}{dx^2} < 0$

$\therefore x=\frac{10\sqrt{3}}{3}\,\, $is a maximum$$

$To find the maximum volume, we substitute this back into the equation$

$V = x (100-x^2)$

$= \frac{10\sqrt{3}}{3} (100- ( \frac{10\sqrt{3}}{3} )^2 )$

$= \frac{2000\sqrt{3} }{9} cm^3$

$\approx 384.9 cm^3$

Can you please tell me what you used to type this out?

kurt.physics · Mar 13, 2009

^^Its called LaTeX. The best internet guide is at About LaTeX

Maximisation/Minimisation Problems (1 Viewer)

acevipa

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