Unit 2_DB MATH133

# Unit 2_DB MATH133 - A= 66 2W^2 A= (66* 33) * 16.5 A=...

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Unit 2 DB Assignment April 1, 2011 The total length of all three walls that John can afford is 66 feet. What is the formula for the area of the garage? A= L * W Choose two dimensions to plug into your formula. P=L+W+W P=L+12+16 W= 12 and W=16 What areas result? A=L*W A= (33 – 2W) * W A= (66 – 2W^2 A= (66 -2 * 12) * 12 A= (66-24) * 12 A= (42) * 12 A= 504 SQ. FT. A= (66-2 *16) * 16 A=(66 -32) * 16 A= (34) * 16 A= 544.5 SQ. FT. Use the vertex formula to find the dimension that produces the maximum area of the garage. VERTEX FORMULA IS: A= 544.5 - 2 * (X – 16 – 5) ^2 W=-b/ (2a) W= -66/ (2*-2) W= -66/ -4 W= 16.5 FT. and L=66-2W L=66 – 2 *(16.5) L=66 – 33 L= 33 Find the maximum area. The Length “L” = 33 sq. ft and the Width “W” = 16.5 sq.ft. A= ( 66 – 2 * W) * W

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Unformatted text preview: A= 66 2W^2 A= (66* 33) * 16.5 A= 33*16.5 A= 544.5 The maximum area is A = 544. 5 sq. ft. How does the length of the kitchen wall affect the structure of the garage? The estimated length of the kitchen wall I calculated supports the newly added walls that complete the garage. If I used any variables above 22, as stated in our assignment, the garage may not have been large enough to accommodate Johns needs. The kitchen walls length is imperative for the overall structure which will be a support for the three added walls, by providing a brace like effect for the entire garage. With the maximum area of 544.5 sq. ft., John will have enough room to park both big cars into his newly built garage....
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## This note was uploaded on 12/10/2011 for the course MATH 133 taught by Professor Hutchins during the Spring '10 term at AIU Online.

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Unit 2_DB MATH133 - A= 66 2W^2 A= (66* 33) * 16.5 A=...

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