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.

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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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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.

 

↑
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
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
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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