Answer to Question #156056 in Trigonometry for anne

Question #156056

A poster is to have an area of 180 in with 1-inch margins at the bottom and sides and a 2-inch margin at the top. What dimensions will give the largest printed area?


1
Expert's answer
2021-01-19T18:13:53-0500

Let the poster is of width "a" and length "b". Then "b=\\frac{180}{a}". Let us sketch the picture:



It follows that the printed part of poster is of width "a-2" and length "b-3". The printed area is equal

"S(a)=(a-2)(\\frac{180}{a}-3)=186-\\frac{360}{a}-3a".


Then "S'(a)=\\frac{360}{a^2}-3". If "S'(a)=0" , then "3a^2=360", and thus "a=\\sqrt{120}=2\\sqrt{30}" (since "a>0"). It follows that "b=\\frac{180}{2\\sqrt{30}}=3\\sqrt{30}". Since "S'(a)=-\\frac{720}{a^3}<0" for "a>0", we conclude that "a=2\\sqrt{30}" is a point of maximum of area function "S(a)". Therefore, the largest printed area for poster of width "2\\sqrt{30}" and length "3\\sqrt{30}."



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS