Answer to Question #154585 in Calculus for Phyroe

Question #154585

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


1
Expert's answer
2021-01-12T01:14:53-0500

Given, hypotenuse = 10 units.

Let Base be b units.

And perpendicular be p units.

Then by pythagoras theorem,

perpendicular2+base2=hypotenuse2perpendicular^2+base^2=hypotenuse^2

p2+b2=102p2+b2=100p^2+b^2=10^2 \\ \Rightarrow p^2+b^2=100

p=100b2\Rightarrow p=\sqrt{100-b^2} ...(i)

Now, area of right triangle =12×base×height=12×p×b=\frac12 \times base \times height=\frac12 \times p \times b

A=12pbA = \frac12p b ...(ii)

Putting value from (i) into (ii),

A=12b100b2A=\frac12 b\sqrt{100-b^2}

On differentiating w.r.t b,

dAdb=12[(b)100b2+b(100b2)]=12[(1)100b2+b×12100b2×(02b)]=12[100b2b2100b2]\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 dAdb=0\frac{dA}{db}=0

12[100b2b2100b2]=0100b2b2100b2=0100b2=b2100b2100b2×100b2=b2(100b2)=b2100=2b2b2=50b=50=52\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)

dAdb=12[(100b2)b2100b2]=12[1002b2100b2]\frac{dA}{db}=\frac12 [\frac{(100-b^2)-b^2}{\sqrt{100-b^2}}]=\frac12 [\frac{100-2b^2}{\sqrt{100-b^2}}]

dAdb=50b2100b2\Rightarrow \frac{dA}{db}=\frac{50-b^2}{\sqrt{100-b^2}}

Again differentiating w.r.t b,

d2Adb2=(100b2)(02b)(50b2)(12100b2)(02b)100b2\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}}

=2b100b2+(b(50b2)100b2)100b2=\frac{-2b\sqrt{100-b^2}+(\frac{b(50-b^2)}{\sqrt{100-b^2}})}{{100-b^2}}

=2b(100b2)+b(50b2)(100b2)3/2=\frac{-2b({100-b^2})+b(50-b^2)}{{(100-b^2)^{3/2}}}

=200b+2b2+50bb3(100b2)3/2=150b+2b2b3(100b2)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,

d2Adb2=150(52)+2(50)(50)3/2(10050)3/2\frac{d^2A}{db^2}=\frac{-150(5\sqrt{2})+2(50)-(50)^{3/2}}{{(100-50)^{3/2}}}

=750(2)+100(50)3/2(50)3/2<0=\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=10050=50=52p=\sqrt{100-50}=\sqrt{50}=5\sqrt{2}

Answer: Other two sides are 52,525\sqrt{2},5\sqrt{2} units.


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