Optimization+Continued

Optimization Continued

-Took up homework from yesterday p. 309-311 #1-24 -New Homework p. 401 #1-24

p. 311 #24

Someone wants to build a rectangular kennel within a right angled triangle (yard). What are the dimensions of the kennel that maximize the possible area.



Step 1: What is the slope of the hypoteneuse?

m= rise/run m= - 4/12 m= -1/3

Step 2: What is the Y-intercept? The y intercept of the line is 4.

Step 3: What is the equation of the hypoteneuse?

y= - 1/3 x + 4

Step 4: What is the equation of the area?

A= xy A= x (- 1/3x+4) A= - 1/3x^2 +4x

Step 5: what is the derivative of the area? (when equal to zero)

A'= -2/3x +4 0= - 2/3x +4 x=6

Step 6: Sub the x value into the original line equation.

y = -1/3x +4 y= -1/3 (6) +4 y=2

Therefore, the dimensions that maximize the area would be 6m and 2m.

ALSO, you could solve this problem by comparing the ratios between y and x using simmilar triangles.

Step 1: find the ratio.

y/4 = x/12

y=1/3x

Step 2: Find an expression for the Area/dimensions of the kennel.

The horizontal value of the kennel would be 12-x. with x being the horizontal dimension of the right angle triangle formed as a result of the kennel being built.

A= xy A=(12-x)y A=(12-x)(1/3x) A=4x-1/3x^2

this gives you the same basic equation as the other method. from here, you would continue the same way. take the derivative, and solve.

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