Find the dimensions of a rectangle with perimeter 108 m whose area is as large as possible. Part 1 of 4 If a rectangle has dimensions x and y, then we must maximize the area A = xy. Since the perimeter is 2x + 2y = 108, then y = 54 - x. Part 2 of 4 We must maximize the area A = xy = x(54 - x) = 54x - x^2, where 0 < x < 54. Part 3 of 4 Solving 0 = A'(x) gives us x =
Added by William M.
Close
Step 1
The dimensions of the rectangle are x=8, y=12. Show more…
Show all steps
Your feedback will help us improve your experience
Andrew Noble and 80 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Find the dimensions of a rectangle with perimeter 104 m whose area is as large as possible. If a rectangle has dimensions x and y, then we must maximize the area A = xy. Since the perimeter is 104, then 2x + 2y = 104. We must maximize the area A = xy. where 0 < x < 52. Solving A(x) gives us A(x) = x(52 - x). Step 1: Since A'(x) = 52 - 2x, we set A'(x) = 0 to find critical points. Step 2: 52 - 2x = 0 2x = 52 x = 26 Step 3: Since A''(x) = -2, which is negative, A(x) has an absolute maximum at x = 26. The other dimension of this rectangle is y = 52 - x = 52 - 26 = 26. Thus, the dimensions of the rectangle with perimeter 104 and maximum area are both 26. Step 4:
Andrew N.
Maximizing Area Find the dimensions of the rectangle with the largest area that can be enclosed on all sides by $L$ meters of fencing.
Applications of the Derivative
Optimization
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD