Consider a rectangle of length L inches and width W inches. Find a formula for the perimeter of the rectangle. Use upper case letters. P = L+L+W+W (b) If the length and width of the rectangle are changing with respect to time, find dP dt . Use dL dt and dW dt and not L ' and W ' . dP dt = 2( dL dt)+2( dW dt) (c) Suppose the length is increasing at 2 inches per hour and the width is decreasing at 3 inches per hour. How fast is the perimeter of the rectangle changing when the length is 40 inches and the width is 104 inches?

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valetta valetta · Answer 1 · 2019-07-22T08:28:14+02:00

Answer:

a) P=2(L+W)

b)[tex]\frac{dp}{dt}=2\frac{dL}{dt}+2\frac{dW}{dt}[/tex]

c)-2 inch/hour

Step-by-step explanation:

given:

length of the rectangle as L inches

width of the rectangle as W inches

a) The perimeter is defined as the measure of the exterior boundaries

therefore, for the rectangle the perimeter 'P' will be

P= length of AB+BC+CD+DA (A,B,C and D are marked on the figure attached)

Now from figure

P= L+W+L+W

OR

=> P=2L+2W .....................(1)

b)now dp/dt can be found as by differentiating the equation (1)

[tex]\frac{dP}{dt}=2(\frac{dL}{dt} )+2(\frac{dW}{dt} )[/tex] .............(2)

c)Now it is given for the part c of the question that

L=40 inches

W=104 inches

dL/dt=2 inches/hour

dW/dt= -3 inches/hour (here the negative sign depicts the decrease in the dimension)

substituting the above values in the equation (2) we get

[tex]\frac{dP}{dt}=2(2)+2(-3)[/tex]

[tex]\frac{dP}{dt}=4-6=-2 inches/hour[/tex]