Question

PLEASE HELP ME! Final t0day! (1)A poster is to have an area of 180in^2with 1 inch margins at the bottom and sides
and a 2 inch margin at the top. What dimension will give the largest printed area?

Answer 1

call length of poster x and width y. then you knowo
\[xy=180\] making
\[y=\frac{180}{x}\]
area of printed part is
\[(x-2)(y-4)=(x-2)(\frac{180}{x}-4)=\frac{4(45-x)(x+2)}{x}\]

Answer 2

This is an optimizastion problem:
Maximize: A = (w-2)(h-3)
Constraint: 180=wh

Answer 3

derivative if
\[\frac{360-4x^2}{x^2}\] set equal zero, get
\[x=3\sqrt{10}\] check my algebra

Answer 4

psycho i right i am wrong. i should learn to read

Answer 5

it says one inch around and two at the top. so equation i had was incorrect.

Answer 6

should be
\[A(x)=(x-2)(\frac{180}{x}-3)=-3x-\frac{360}{x}+186\]

Answer 7

solve the constrint for h and plug it into the equation for the printed area.

A(w)=(w-2)(180/w - 3)

Answer 8

\[A'(x)=-3+\frac{360}{x^2}\] set equal zero get
\[3x^2=360\]
\[x=2\sqrt{30}\]

Answer 9

\[A(w)=186-3w-\frac{360}{w}\]
\[ A'(w)=-3+\frac{360}{w^2}\]

From the derivative you can find three critical points w=0, \[w=\pm \sqrt{\frac{360}{3}} = \pm10.9544512\]
since we are working with dimensions we know w>0, so only 10.95445 makes sense.

Answer 10

So we know that the maximum printed area will be at w=10.95445

Answer 11

h= 180/10.95445 = 16.4316767