# Optimization find dimensions problem

• Asphyxiated
In summary, the conversation discusses finding the dimensions of a poster with the smallest area given fixed top and bottom margins of 6 cm and side margins of 4 cm. The equation (a-6)(b-4)=384 is used to solve for a and substitute it into the area formula A=ab, resulting in the equation A(b)=\frac{360b+6b^{2}}{b-4}. After taking the derivative and solving for b, the incorrect solution of b=20 and a=30 is obtained. The mistake is identified as not taking into account the two horizontal and two vertical margins in the initial equation.

## 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?

Asphyxiated said:

## 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.

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