bbb911
Answered

Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

The length and width of a rectangle have a sum of 90. What dimensions give the maximum area?

Sagot :

crisforp
2( l + w) = 90, where l, w are dimensions of the rectangle => l + w = 45;
The maximum area is obtain when l = w;
Then l = w = 45/2 = 22.5
[tex]w-width\\l-length\\\\2w+2l=90\\2l=90-2w\ \ \ \ /:2\\l=45-w\\\\A=wl\to A=w(45-w)=45w-w^2[/tex]

[tex]Area\ of\ a\ rectangle\ is\ a\ square\ function.\\The\ maximum\ area\ is\ equal\ to\ the\ y-coordinate\ of\ vertex\\and\ "w"\ is\ equal\ to\ the\ x-coordinate\ of\ vertex.\\\\A(w)=45w-w^2\\\\a=-1;\ b=45;\ c=0\\\\x-coordinate\ of\ vertex:\frac{-b}{2a}\\\\w=\frac{-45}{2\cdot(-1)}=\frac{-45}{-2}=22.5\\\\l=45-22.5=22.5\\\\Solution:22.5\times22.5\ \ (This\ rectangle\ is\ a\ square).[/tex]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.