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Given the nonnegative real number x, YZ satisfies X-1 / 2 = 2-y / 3 = Z-3 / 4, w = 3x + 4Y + 5Z. Find the maximum and minimum value of wde
Let X-1 / 2 = 2-y / 3 = Z-3 / 4 = K
x=k+1/2
y=6-3k
z=k+3/4
W=3(k+1/2)+4(6-3k)+5(k+3/4)
=-4k+117/4
When k = 0, the maximum W is 117 / 4
When y = 0, k = 2, W is 85 / 4
Given that P and Q are circle x ^ 2 + (Y-2) ^ 2 = 1 △ 4 and ellipse x ^ 2 △ 4 + y ^ 2 = 1 respectively, find the maximum value of PQ
To maximize PQ, two points are farthest apart
Then p must be the highest point, i.e. P (0,5 / 2)
Q must be the lowest point, i.e. Q (0, - 1)
So PQ = √ [(0-0) &# 178; + (5 / 2 + 1) &# 178;]
=7/2
Let p be a point on the ellipse x ^ / 25 + y ^ / 9 = 1 and Q R be a point on the circle (x + 4) ^ + y ^ = 1 / 4 and (x-4) ^ + y ^ = 1 / 4 respectively, then what is the minimum value of PQ + PR,
You need to draw
Let the left and right focus of ellipse be F1 and F2
From the difference between the two sides of the triangle is greater than the third side, we know that the minimum PQ is pf1-r
Similarly, the minimum PR is pf2-r
The minimum of PQ + PR is Pf1 + pf2-2r = 2a-2r = 10-1 = 9
Given that point a (0,1) is a point on the ellipse x ^ 2 + 4Y ^ 2 = 4, and P is a moving point on the ellipse, when the length of chord AP is the largest, the coordinate of point P is?
The parameter equation of ellipse x = 2cos @, y = sin @, distance formula from point to point, y ^ 2=（ 2cos@-0 )^2+( sin@-1 ）^2. Expand the equation and change cos @ into sin @. The axis of symmetry is in the definition domain of sin @, where y value is the largest, that is, the distance is the farthest, sin @ = - 1 / 3. Just substitute it into the previous parameter equation
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- 1. Find the maximum length of the chord AP of a point a (1,0) on the ellipse x ^ 2 + 4Y ^ 2 = 4 Sorry, it's (0, 1) wrong. Sorry
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