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



Answer :

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]