Answer:
the dimension of the poster = 90 cm length and 60 cm width i.e 90 cm by 60 cm.
Step-by-step explanation:
From the given question.
Let p be the length of the of the printed material
Let q be the width of the of the printed material
Therefore pq = 2400 cm ²
q = [tex]\dfrac{2400 \ cm^2}{p}[/tex]
To find the dimensions of the poster; we have:
the length of the poster to be p+30 and the width to be [tex]\dfrac{2400 \ cm^2}{p} + 20[/tex]
The area of the printed material can now be: [tex]A = (p+30)(\dfrac{2400 }{p} + 20)[/tex]
=[tex]2400 +20 p +\dfrac{72000}{p}+600[/tex]
Let differentiate with respect to p; we have
[tex]\dfrac{dA}{dp}= 20 - \dfrac{72000}{p^3}[/tex]
Also;
[tex]\dfrac{d^2A}{dp^2}= \dfrac{144000}{p^3}[/tex]
For the smallest area [tex]\dfrac{dA}{dp }=0[/tex]
[tex]20 - \dfrac{72000}{p^2}=0[/tex]
[tex]p^2 = \dfrac{72000}{20}[/tex]
p² = 3600
p =√3600
p = 60
Since p = 60 ; replace p = 60 in the expression q = [tex]\dfrac{2400 \ cm^2}{p}[/tex] to solve for q;
q = [tex]\dfrac{2400 \ cm^2}{p}[/tex]
q = [tex]\dfrac{2400 \ cm^2}{60}[/tex]
q = 40
Thus; the printed material has the length of 60 cm and the width of 40cm
the length of the poster = p+30 = 60 +30 = 90 cm
the width of the poster = [tex]\dfrac{2400 \ cm^2}{p} + 20[/tex] = [tex]\dfrac{2400 \ cm^2}{60} + 20[/tex] = 40 + 20 = 60
Hence; the dimension of the poster = 90 cm length and 60 cm width i.e 90 cm by 60 cm.
