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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Medicine In order to treat a certain bacterial infection, a combination of two drugs is being tested. Studies have shown that the duration of the infection in laboratory tests can be modeled by

D ( x , y ) = x 2 + 2 y 2 18 x 24 y + 2 x y + 120 where x is the dosage (in hundreds of milligrams) of the first drug and y is the dosage (in hundreds of milligrams) of the second drug. Find the amount of each drug necessary to minimize the duration of the infection.

To determine

To calculate: The amount of each drugs necessary to minimize the duration of the infection in laboratory test is given as, D(x,y)=x2+2y218x24y+2xy+120.

Explanation

Given Information:

Duration at infection in laboratory test is given as,

D(x,y)=x2+2y218x24y+2xy+120

Where x is the dosage of the first drug (in hundreds of milligrams),

And y is the dosage of the second drug (in hundreds of milligrams).

Formula used:

Two minimize the function with two variable following procedure can be followed,

Step 1: Write the primary equation for the function to be minimize.

Step 3: Find first partial derivative of primary equation and equate to zero, to find critical point

as,

Dx(x,y)=0 and Dy(x,y)=0

Step 4: Find double partial derivative with respect to all variables separately at different critical

points.

Suppose (a,b) is critical point, then find double partial derivative like-

Dxx(a,b), Dyy(a,b), Dxy(a,b) and Dyx(a,b)

Then to test the function is minimum, maximum and having saddle point consider the

secondary equation,

d=Dxx(a,b)×Dyy(a,b)[Dxy(a,b)]2

(1) If d>0 and Dxx(a,b)>0 then function D has a relative minimum at (a,b).

(2) If d<0 and Dxx(a,b)<0 then function D has a relative maximum at (a,b).

(3) If d<0 then [(a,b),D(x,y)] is a saddle point.

(4) The text gives no information when d=0

Calculation:

Considering the primary equation,

D(x,y)=x2+2y218x24y+2xy+120

Differentiating primary equation with respect to x keeping y as constant.

Dx(x,y)=dD(x,y)dxDx(x,y)=2x+0180+2yDx(x,y)=2x+2y18

Differentiating primary equation with respect to y keeping x as constant.

Dy(x,y)=dD(x,y)dyDy(x,y)=ddy(x2+2y218x24y+2xy+120)Dy(x,y)=4y24+2x

Equating Dx(x,y) with Dy(x,y) to zero for critical points: -

Dx(x,y)=02x+2y18=0

And

Dy(x,y)=04y24+2x=0

Now solving Dx(x,y)=0 and Dy(x,y)=0 for getting (x,y) critical points: -

Dx(x,y)Dy(x,y)=0(2x+2y18)(4y24+2x)=0(2y+6)=0y=3

Substitute 3 for y in Dx(x,y)=2x+2y18=0,

Dx(x,y)=02x+2y18=02x+2×318=0x=6

Thus, critical point (a,b)(6,3)

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