Optimization find dimensions problem

[Asphyxiated]

Homework Statement

The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm. If the area of printed material on the poster is fixed at 384 cm², find the dimensions of the poster with the smallest area.

Homework Equations

Area of a rectangle is length times width

The Attempt at a Solution

So I assumed that the 384 cm² formed a constraint like so:

(a-6)(b-4)=384

and then using this equation I solved for a to be able to substitute it into the area formula:

ab-4a-6b+24=384

ab-4a=360+6b

a(b-4)=360+6b

a= \frac {360+6b}{b-4}

so then i can substitute a in A=ab so that A will be a function of b:

A=ab

A=\frac{360+6b}{b-4} b

A(b)=\frac{360b+6b^{2}}{b-4}

ok, so then i take the derivative of this A(b) now which turns out to be:

A'(b)=\frac {6b^{2}-48b-1440}{(b-4)^{2}}

then i take the top, complete the square and then set it equal to zero:

6(b-4)^{2}-1536 = 0

6(b-4)^{2} = 1536

(b-4)^{2} = 256

b-4 = \sqrt{256}= \pm 16

b= \pm 16 +4

I chose to leave out the negative square root as it makes no sense when we are talking about lenth, so with b=20, take that back and plug it into the a=b equation:

a = \frac {360+6(20)} {20-4} = 30

and I get a = 20 and b=30 and if you take off the margins and multiply them at 16*24 you get 384 cm², but alas, this is not correct. Where did I go wrong at?

 

[Mark44]

Homework Statement

The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm. If the area of printed material on the poster is fixed at 384 cm², find the dimensions of the poster with the smallest area.

Homework Equations

Area of a rectangle is length times width

The Attempt at a Solution

So I assumed that the 384 cm² formed a constraint like so:

(a-6)(b-4)=384

and then using this equation I solved for a to be able to substitute it into the area formula:

ab-4a-6b+24=384

ab-4a=360+6b

a(b-4)=360+6b

a= \frac {360+6b}{b-4}

so then i can substitute a in A=ab so that A will be a function of b:

A=ab

A=\frac{360+6b}{b-4} b

A(b)=\frac{360b+6b^{2}}{b-4}

ok, so then i take the derivative of this A(b) now which turns out to be:

A'(b)=\frac {6b^{2}-48b-1440}{(b-4)^{2}}

then i take the top, complete the square and then set it equal to zero:

6(b-4)^{2}-1536 = 0

6(b-4)^{2} = 1536

(b-4)^{2} = 256

b-4 = \sqrt{256}= \pm 16

b= \pm 16 +4

I chose to leave out the negative square root as it makes no sense when we are talking about lenth, so with b=20, take that back and plug it into the a=b equation:

a = \frac {360+6(20)} {20-4} = 30

and I get a = 20 and b=30 and if you take off the margins and multiply them at 16*24 you get 384 cm², but alas, this is not correct. Where did I go wrong at?

Right near the beginning (emphasis added).

The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm.

Your equation (a - 6)(a - 4) = 384 doesn't take into account the two horizontal and two vertical margins.

 

[Asphyxiated]

thanks, came out right this time, now i got another question posted though... lol