i have two answers but am not sure which one is the correct answer and i need help to solve it i add two pic add help me solve it Problem: A piece of wire, 400 cm long, is to be bent into an isosceles triangle. What should the dimensions of the triangle be in order to maximize its area? Use the First and Second Derivative Tests (derivatives, calculations, and tables for FDT and SDT) to check that your answer(s) is (are) indeed, for the maximum. Determine the maximum value for the area of this triangle. Round you answer(s) to the nearest tenth.
i have two answers but am not sure which one is the correct answer and i need help to solve it i add two pic add help me solve it Problem: A piece of wire, 400 cm long, is to be bent into an isosceles triangle. What should the dimensions of the triangle be in order to maximize its area? Use the First and Second Derivative Tests (derivatives, calculations, and tables for FDT and SDT) to check that your answer(s) is (are) indeed, for the maximum. Determine the maximum value for the area of this triangle. Round you answer(s) to the nearest tenth.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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i have two answers but am not sure which one is the correct answer and i need help to solve it i add two pic add help me solve it
Problem: A piece of wire, 400 cm long, is to be bent into an isosceles triangle. What should the dimensions of the triangle be in order to maximize its area? Use the First and Second Derivative Tests (derivatives, calculations, and tables for FDT and SDT) to check that your answer(s) is (are) indeed, for the maximum. Determine the maximum value for the area of this triangle. Round you answer(s) to the nearest tenth.
Transcribed Image Text:↑
Given that a wire is 400 cm.
So, a + b + c = 370
Since triangle is isosceles triangle
So, let's assume that a=b
So, 2a + b = 400 b = 400 - 2a
The area of an isosceles triangle is,
bh ¹ × 400 - 2a√370² +4a² - 1600a - a²
(200-a)1600² + 3a² - 1600a
=
=
-
Step 2: Step 2
Take the derivative,
(200-a)400² + 3a² - 1600a¹3a - 800 -√400² + 3a² - 1600a = 0 ⇒ (200-a) (3a - 800) - 400² -3a² + 1600a = 0⇒200 × 3a
·
So, the sides are, 154, 154, 92
Solution
So, the sides are, 154, 154, 92
![A² = 10 [-2] √₂-100 + 10 (400-2x)
2√2-100
A² = 0
-20 √√√2-100 + 5 (400 - 2a)
2-100
=) -20 (9-100) + 5 (400-2x) = 0
=) - 202 + 2000 + 2000 - 10α = 0
=)
11
A = -10
21-100
30x + 4000
.. X =
2 = 400
3
+
552)
x-100
= 0
~ 188.3
(-2)_ 5 (400-2x)
2(x-100) ³/2
• 1" (400/3) 20
so Areen is max at 2 = 400
3
:. Sides are 123.3, 133.2, 133.3 Ans
~133-3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdc8dfbd3-eff9-44bf-a936-fc352fedaeee%2Fc12eed31-aa15-4729-805b-b8e29bb88f35%2Fd1z1mag_processed.png&w=3840&q=75)
Transcribed Image Text:A² = 10 [-2] √₂-100 + 10 (400-2x)
2√2-100
A² = 0
-20 √√√2-100 + 5 (400 - 2a)
2-100
=) -20 (9-100) + 5 (400-2x) = 0
=) - 202 + 2000 + 2000 - 10α = 0
=)
11
A = -10
21-100
30x + 4000
.. X =
2 = 400
3
+
552)
x-100
= 0
~ 188.3
(-2)_ 5 (400-2x)
2(x-100) ³/2
• 1" (400/3) 20
so Areen is max at 2 = 400
3
:. Sides are 123.3, 133.2, 133.3 Ans
~133-3
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