Optimization find dimensions problem

  • #1
264
0

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:

[tex] (a-6)(b-4)=384[/tex]

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

[tex] ab-4a-6b+24=384 [/tex]

[tex] ab-4a=360+6b [/tex]

[tex] a(b-4)=360+6b [/tex]

[tex] a= \frac {360+6b}{b-4} [/tex]

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

[tex] A=ab [/tex]

[tex] A=\frac{360+6b}{b-4} b [/tex]

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

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

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

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

[tex] 6(b-4)^{2}-1536 = 0 [/tex]

[tex] 6(b-4)^{2} = 1536 [/tex]

[tex] (b-4)^{2} = 256 [/tex]

[tex] b-4 = \sqrt{256}= \pm 16 [/tex]

[tex] b= \pm 16 +4 [/tex]

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:

[tex] a = \frac {360+6(20)} {20-4} = 30 [/tex]

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?
 

Answers and Replies

  • #2
34,892
6,630

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:

[tex] (a-6)(b-4)=384[/tex]

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

[tex] ab-4a-6b+24=384 [/tex]

[tex] ab-4a=360+6b [/tex]

[tex] a(b-4)=360+6b [/tex]

[tex] a= \frac {360+6b}{b-4} [/tex]

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

[tex] A=ab [/tex]

[tex] A=\frac{360+6b}{b-4} b [/tex]

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

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

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

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

[tex] 6(b-4)^{2}-1536 = 0 [/tex]

[tex] 6(b-4)^{2} = 1536 [/tex]

[tex] (b-4)^{2} = 256 [/tex]

[tex] b-4 = \sqrt{256}= \pm 16 [/tex]

[tex] b= \pm 16 +4 [/tex]

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:

[tex] a = \frac {360+6(20)} {20-4} = 30 [/tex]

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.
 
  • #3
264
0
thanks, came out right this time, now i got another question posted though.... lol
 

Related Threads on Optimization find dimensions problem

Replies
1
Views
557
Replies
8
Views
3K
Replies
2
Views
1K
Replies
7
Views
4K
Replies
2
Views
2K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
9
Views
3K
Top