Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
Find the two numbers whose sum is 15 and the square of one
multiplied by the cube of the other is maximum.
Hi Shaleen.
Lets consider the two numbers to be x,y (both greater than 0).
So x+y = 15
Now to maximise T = x^2 * y^3
So T = y^3 (15-y)^2 = y^3 { 225 + y^2 - 30y)
So T = 225y^3 + y^5 - 30y^4.
Now maximise this using dT/dy = 0.
Clearly y=0 will not maximise "T".
Hence y(2y-30) + 3 { 225 + y^2 - 30y } = 0.
Solve this QE, for y value (for which the maximum occurs).
Hope that helps.
Best Regards,
Ashwin (IIT Madras).
Algorithm:
First form a equations i. x+y=15 ----(1) and (x^2)(y^3) is maximum
Now take the value of x from equation (1).
=((15-y)2)(y3)
=(225+y2-30y)(y3)
=(225y3 +y5 -30y4 )
Now differentiate in terms of y and putting it equal to 0 which is the slope of the graph at the maxima or minima
0=675y2+5y4-120y3
0=5y2(y2-24y+135)
y2-24y+135=0
The above equation are satisfied only at 15,9
At 15,0 it is at the minima, So y=9 and x=6 must be the correct answers
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !