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Firefox File Edit View History Bookmarks Tools Window Help WA 10 Math 1325 Homework s x M Mathway l Math Problem s x C chegg study l Guided solu x sy tangentofox-1'2, -9.2)-- x Search webassign, net /webl Student/Assignment-Responses/last?dep 15484216 Most Visited Q cl if Facebook M gmail GP MyLab M Inbox jolmsteado.- M Mathway Y WebAssign wolfram Lt kickass HH careers Assign men Your last submission is used for your score. 10 0/10 points I Previous Answers LarApCalc10 3.4.003 Find the length and width of a rectangle that has the given area and a minimum perimeter. Area: 64 square feet X feet (smaller value) length X feet (larger value) width Need Help? Read It Talk to a Tutor Submit Answer save Progress Practice Another Version View Previous Question Question 10 of 10 Home My Assignments Extension Request WobAssign 4.0 1997-2017 Advanood Instruotional Syatoma, Inc. All rights reoonvod. 52% Mon 3:03 AM QA Sy symolab My Notes Ask Your Teacher

Explanation / Answer

Area = L*B =64
That means B=64/L

Perimeter, P=L+B
or
P=L+(64/L)
For the perimeter to be minimum the differential of P with respect to L must be zero
Thus, dP/dL=0.
d/dL(L+64/L)=0
1-64/(L*L)=0
1=64/(L*L)
L*L=64
L=square root of (64) = 8 ft
Since
B=64/L
B = 64/ 8 = 8 ft

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