Question

If the length of the hypotenuse of a right triangle is 10, find the length of the other sides when the area is maximum.

Answer 1

Given, hypotenuse = 10 units.

Let Base be b units.

And perpendicular be p units.

Then by pythagoras theorem,

$perpendicular^2+base^2=hypotenuse^2$

$p^2+b^2=10^2 \\ \Rightarrow p^2+b^2=100$

$\Rightarrow p=\sqrt{100-b^2}$ ...(i)

Now, area of right triangle $=\frac12 \times base \times height=\frac12 \times p \times b$

$A = \frac12p b$ ...(ii)

Putting value from (i) into (ii),

$A=\frac12 b\sqrt{100-b^2}$

On differentiating w.r.t b,

$\frac{dA}{db}=\frac12 [(b)'\sqrt{100-b^2}+b(\sqrt{100-b^2})'] \\=\frac12 [(1)\sqrt{100-b^2}+b\times \frac{1}{2\sqrt{100-b^2}}\times(0-2b)] \\=\frac12 [\sqrt{100-b^2}- \frac{b^2}{\sqrt{100-b^2}}]$ ...(iii)

Now, put $\frac{dA}{db}=0$

$\frac12 [\sqrt{100-b^2}- \frac{b^2}{\sqrt{100-b^2}}]=0 \\ \Rightarrow \sqrt{100-b^2}- \frac{b^2}{\sqrt{100-b^2}}=0 \\ \Rightarrow \sqrt{100-b^2}=\frac{b^2}{\sqrt{100-b^2}} \\ \Rightarrow \sqrt{100-b^2} \times \sqrt{100-b^2}=b^2 \\ \Rightarrow (100-b^2)=b^2 \\ \Rightarrow 100=2b^2 \\ \Rightarrow b^2=50 \\ \Rightarrow b=\sqrt{50}=5\sqrt{2}$ ....(iv)

Consider (iii)

$\frac{dA}{db}=\frac12 [\frac{(100-b^2)-b^2}{\sqrt{100-b^2}}]=\frac12 [\frac{100-2b^2}{\sqrt{100-b^2}}]$

$\Rightarrow \frac{dA}{db}=\frac{50-b^2}{\sqrt{100-b^2}}$

Again differentiating w.r.t b,

$\frac{d^2A}{db^2}=\frac{(\sqrt{100-b^2})(0-2b)-(50-b^2)(\frac{1}{2\sqrt{100-b^2}})(0-2b)}{{100-b^2}}$

$=\frac{-2b\sqrt{100-b^2}+(\frac{b(50-b^2)}{\sqrt{100-b^2}})}{{100-b^2}}$

$=\frac{-2b({100-b^2})+b(50-b^2)}{{(100-b^2)^{3/2}}}$

$=\frac{-200b+2b^2+50b-b^3}{{(100-b^2)^{3/2}}}=\frac{-150b+2b^2-b^3}{{(100-b^2)^{3/2}}}$

Putting value from (iv) here,

$\frac{d^2A}{db^2}=\frac{-150(5\sqrt{2})+2(50)-(50)^{3/2}}{{(100-50)^{3/2}}}$

$=\frac{-750(\sqrt{2})+100-(50)^{3/2}}{{(50)^{3/2}}} <0$

So, area is maximum.

Now, put value of b from (iv) in (i),

$p=\sqrt{100-50}=\sqrt{50}=5\sqrt{2}$

Answer: Other two sides are $5\sqrt{2},5\sqrt{2}$ units.