The bottom of the legs of a three legged table are the vertices of an isoceles triangle with sides 5,5 and 6. The legs are to be braced at the bottom by three wires in the shape of a Y. The minimum length of the wire needed for this purpose, is

Question

The bottom of the legs of a three legged table are the vertices of an isoceles triangle with sides 5,5 and 6. The legs are to be braced at the bottom by three wires in the shape of a Y. The minimum length of the wire needed for this purpose, is (A) 4+3 √3 (B) 10 (C) 3+4 √3 (D) 1+6 √2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/aaf4dc2a943eae792d45e6456db7d52c-.png" /> L=4-x+2 √9+x2 where 0 ≤ x ≤ 4 (d L/d x)=(2 x/√9+x2)-1=0 4 x2=9+x2 x=√3 now L (0)= L (4)=10 L (√3)=4-√3+2 ⋅ 2 √3=4+3 √3

Q. The bottom of the legs of a three legged table are the vertices of an isoceles triangle with sides 5,5 and 6. The legs are to be braced at the bottom by three wires in the shape of a Y. The minimum length of the wire needed for this purpose, is

$4 + 33$

10

$3 + 43$

$1 + 62$

$L = 4 - x + 2 9 + x^{2}$ where $0 \leq x \leq 4$
$\frac{d L}{d x} = \frac{2 x}{9 + x ^{2}} - 1 = 0$
$4 x^{2} = 9 + x^{2}$
$x = 3$
now $L (0) = L (4) = 10$
$L (3) = 4 - 3 + 2 \cdot 23 = 4 + 33$

Q. The bottom of the legs of a three legged table are the vertices of an isoceles triangle with sides 5,5 and 6. The legs are to be braced at the bottom by three wires in the shape of a Y. The minimum length of the wire needed for this purpose, is

4+33​

10

3+43​

1+62​

L=4−x+29+x2​ where 0≤x≤4 dxdL​=9+x2​2x​−1=0 4x2=9+x2 x=3​ now L(0)=L(4)=10 L(3​)=4−3​+2⋅23​=4+33​

$4 + 33$

$3 + 43$

$1 + 62$

$L = 4 - x + 2 9 + x^{2}$ where $0 \leq x \leq 4$
$\frac{d L}{d x} = \frac{2 x}{9 + x ^{2}} - 1 = 0$
$4 x^{2} = 9 + x^{2}$
$x = 3$
now $L (0) = L (4) = 10$
$L (3) = 4 - 3 + 2 \cdot 23 = 4 + 33$